TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

I.
Question 1.
Find the set of vaines of x for which the binomial expansions of the following are valid.
i) (2 + 3x)\(\frac{-2}{3}\)
ii) (5 + x)\(\frac{3}{2}\)
iii) (7 + 3x)-5
iv) (4 – \(\frac{x}{3}\))\(\frac{-1}{2}\)
Solution:
We know that
(2 + 3x)\(\frac{-2}{3}\) = 2\(\frac{-2}{3}\) (1 + \(\frac{3}{2}\)x)\(\frac{-2}{3}\)
The expansion is valid only when |\(\frac{3}{2}\) x| < 1
⇒ |x| < \(\frac{2}{3}\)
⇒ x ∈ \(\left(\frac{-2}{3}, \frac{2}{3}\right)\).

ii) We know that (5 + x)3/2
= 53/2 (1 + \(\frac{x}{5}\))3/2
The expansion is valid only when
|\(\frac{x}{5}\)| < 1
⇒ |x| < 5
⇒ x ∈ (- 5, 5).

iii) We know that
(7 + 3x)-5 = 7-5 (1 + \(\frac{3 x}{7}\))-5
The expansion is valid only when |\(\frac{3 x}{7}\)| < 1
⇒ |x| < \(\frac{7}{3}\)
⇒ x ∈ \(\left(\frac{-7}{3}, \frac{7}{3}\right)\)

iii) We know that (4 – \(\frac{x}{3}\))\(\frac{-1}{2}\)
= \(4^{\frac{-1}{2}}\left(1-\frac{x}{12}\right)^{\frac{-1}{2}}\)
The expansion is valid only when |\(\frac{x}{12}\)| < 1
⇒ |x| < 12
⇒ x ∈ (- 12, 12).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 2.
i) 6th term of (1 + \(\frac{x}{2}\))-5
ii) 7th term of (1 – \(\frac{x^2}{3}\))-4
iii) 10th term of (3 – 4x)\(\frac{-2}{3}\)
iv) 5th term of (7 + \(\frac{8 y}{3}\))\(\frac{7}{4}\)
Solution:
i) General term in the expansion of (1 + x)-n is
Tr+1 = (- 1)r . n+r-1Cr . xr
∴ 6th term of (1 + x)-5 is
T6 = (- 1)5 . 9C5 (\(\frac{x}{2}\))-5 (∵ n = r = 5)
= \(\frac{-63}{16}\) x5.

ii) General term in the expansion of (1 + x)-n is
Tr+1 = (- 1)r . n+r-1Cr . xr
7th term in the eapansion of (1 – \(\frac{x^2}{3}\))-4 is
T7 = \({ }^9 C_6\left(\frac{x^2}{3}\right)^6\) (∵ n – 4; r = 6)
= \(\frac{28}{243}\) x12

iii) We know that
(3 – 4x)\(\frac{-2}{3}\) = \(3^{\frac{-2}{3}}\left(1-\frac{4 x}{3}\right)^{\frac{-2}{3}}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 1

iv) We know that (7 + \(\frac{8 y}{3}\))\(\frac{7}{4}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 2

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 3.
Write down the first 3 terms In the expansion of
i) (3 +5x)\(\frac{-7}{3}\)
ii) (1 + 4x)-4
iii) (8 – 5x)\(\frac{2}{3}\)
iv) (2 – 7x)\(\frac{-3}{4}\)
Solution:
i) \((3+5 x)^{\frac{-7}{3}}=3^{\frac{-7}{3}}\left(1+\frac{5 x}{3}\right)^{\frac{-7}{3}}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 3

ii) We know that
(1 + x)-n = 1 – nx + \(\frac{\mathrm{n}(\mathrm{n}+1)}{2 !}\) x2 – ………….
∴ (1 + 4x)-4 = 1 – 4 (4x) + \(\frac{4(5)}{2}\) (4x)2 – …………..
= 1 – 16x + 160x2 – ……………
∴ The first 3 terms are 1, – 16x, 160x2.

iii) (8 – 5x)\(\frac{2}{3}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 4

iv) \((2-7 x)^{\frac{-3}{4}}=2^{\frac{-3}{4}}\left[1-\frac{7 x}{2}\right]^{\frac{-3}{4}}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 5

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
Find the general term ((r + i)th term) in the expansion of
i) (4 + 5x)\(\frac{-3}{2}\)
ii) (1 – \(\frac{5 x}{3}\))-3
iii) (1 + \(\frac{4 x}{5}\))\(\frac{5}{2}\)
iv) (3 – \(\frac{5 x}{4}\))\(\frac{-1}{2}\)
Solution:
i) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 6

ii) General term in the expansion (1 – x)-n is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 7

iii) General term in the expansion of (1 + X)p/q is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 8

iv) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 9

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

II.
Question 1.
Find the coefficient of x10 in the expansion of \(\frac{1+2 x}{(1-2 x)^2}\).
Solution:
We know that
\(\frac{1+2 x}{(1-2 x)^2}\) = (1 + 2x) (1 – 2x)-2
= (1 + 2x) [1 + 2(2x) + 3 (2x)2 + 4 (2x)3 + …………. + 10 (2x)9 + 11 (2x)10 + …………]
∴ The coefficient of x10 in \(\frac{1+2 x}{(1-2 x)^2}\) is
= 11 (210) + 10 (2) (29)
= 210 (11 + 10)
= 21 × 210.

Question 2.
Find the coefficient of x4 in the expansion of (1 – 4x)-3/5.
Solution:
General term in the expansion of (1 – X)\(\frac{-p}{q}\) is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 10

Question 3.
i) Find the coefficient of x5
ii) Find the coefficient of x8 in (1 nt)2
iii) Find the coefficient of x in (2+3×9
Solution:
i) We know that
\(\frac{(1-3 x)^2}{(3-x)^{\frac{3}{2}}}=(1-3 x)^2\left[3-\left.x\right|^{\frac{-3}{2}}\right.\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 11

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 12

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

ii) We know that \(\frac{(1+x)^2}{\left(1-\frac{2}{3} x\right)^3}=(1+x)^2\left[1-\frac{2}{3} x\right]^{-3}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 13

iii) We know that
\(\frac{(2+3 x)^3}{(1-3 x)^4}\) = (2 + 3x)3 [1 – 3x]-4
= (8 + 36x + 54x2 + 27x3) [1 + 4C (3x) + 5C2 (3x)2 + 6C3 (3x)3 + ……….. + (r+3)Cr (3x)r + …………..]
Clearly the coefficient of xr in above expansion is
8 (r+3)Cr 3r + 36 (r+2)Cr-1 3r-1 + 54 (r+1)Cr-2 3r-2 + 27 rCr-3 3r-2
for coefficient of x7, put r = 7
∴The coefficient of x7 in \(\frac{(2 x+3)^3}{(1-3 x)^4}\) is
8 10C7 37 + 36 9C6 36 + 54 8C5 35 + 27 7C4 34 = 8 10C3 37 + 36 9C3 36 + 54 8C3 35 + 27 7C3 34.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
Find the coefficient of x3 in the expansion of \(\frac{\left(1+3 x^2\right)^{\frac{3}{2}}}{(3+4 x)^{\frac{1}{3}}}\).
Solution:
\(\frac{\left(1+3 x^2\right)^{\frac{3}{2}}}{(3+4 x)^{\frac{1}{3}}}\) = (1 + 3x)3/2 (3 + 4x)-1/3

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 14

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

III.
Question 1.
Find the sum of the infinite series:
i) 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}\) + ………….
ii) 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + ………….
iii) \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}\) + …………
iv) \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\) – …………..
Solution:
i) Let y = 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}\) + ………….

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 15

ii) Let y = – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + ………….

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 16

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

iii) \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}\) + …………

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 17

iv) Let y = \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\) – …………..

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 18

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 2.
If t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……….. ∞, then prove that 9t = 16.
Solution:
Given
t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……….. ∞

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 19

Question 3.
If x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\) then prove that 9x2 + 24x = 11.
Solution:
Given x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 20

⇒ \(\frac{4}{3}\) + x = √3
⇒ 4 + 3x = 3√3
⇒ 16 + 9x2 + 24x = 27
⇒ 9x2 + 24x = 11.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
If x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots \ldots\) then find the value of x2 + 4x.
Solution:
x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots \ldots\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 21

∴ (1) ⇒ 2 + x = \(\left(1-\frac{2}{3}\right)^{-\frac{3}{2}}\)
⇒ 2 + x = \(\left(\frac{1}{3}\right)^{\frac{-3}{2}}\)
⇒ 2 + x = √27
⇒ x2 + 4x = 23.

Question 5.
Find the sum to infinite terms of the series \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\)
Solution:
Let y = \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 22

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 6.
Show that for any non-zero rational number x.
\(1+\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots .\) = \(1+\frac{x}{3}+\frac{x(x+1)}{3 \cdot 6}+\frac{x(x+1)(x+2)}{3 \cdot 6 \cdot 9}+\ldots .\)
Solution:
L.H.S = \(1+\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots .\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 23

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Exercise 7(c)

I. Find the values of the following integrals.

Question 1.
\(\int_0^{\frac{\pi}{2}} \sin ^{10} x d x\) (May ’06; Mar. ’03)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q1

Question 2.
\(\int_0^{\frac{\pi}{2}} \cos ^{11} x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 3.
\(\int_0^{\frac{\pi}{2}} \cos ^7 x \sin ^2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q3

Question 4.
\(\int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^4 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q4

Question 5.
\(\int_0^\pi \sin ^3 x \cos ^6 x d x\)
Solution:
We have \(\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\)
if f(2a – x) = f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q5

Question 6.
\(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\)
Solution:
Take f(x) = sin2x cos4x
Then f(π – x) = sin2(π – x) cos4(π – x)
= sin2x cos4x
= f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6.1

Question 7.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta \cos ^7 \theta d \theta\)
Solution:
f(θ) = sin2θ cos7θ
f(-θ) = sin2(-θ) cos7(-θ)
= sin2θ cos7θ
= f(θ); and f is even
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q7

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 8.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta\)
Solution:
Let f(θ) = sin3θ cos3θ
∴ f(-θ) = sin3(-θ) cos3(-θ)
= -sin3θ cos3θ
= -f(θ)
∴ f is an odd function.
Hence \(\int_{-a}^a f(x) d x=0\) when f is odd.
∴ \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta=0\)

Question 9.
\(\int_0^a x\left(a^2-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Take x = a sin θ then dx = a cos θ dθ
Upper limit when x = a is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9
Let cos θ = t then -sin θ dθ = dt
Upper limit when θ = \(\frac{\pi}{2}\) is t = 0
Lower limit when θ = 0 is t = 1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9.1

Question 10.
\(\int_0^2 x^{\frac{3}{2}} \sqrt{2-x} d x\)
Solution:
Take x = 2 sin2θ, then dx = 4 sin θ cos θ dθ
Upper limit when x = 2 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10.1

II. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^5(1-x)^{\frac{5}{2}} d x\)
Solution:
Let x = sin2θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 2.
\(\int_0^4\left(16-x^2\right)^{\frac{5}{2}} d x\)
Solution:
Let x = 4 sin θ then dx = 4 cos θ dθ
Upper limit when x = 4 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q2

Question 3.
\(\int_{-3}^3\left(9-x^2\right)^{\frac{3}{2}} x d x\)
Solution:
Let x = 3 sin θ then dx = 3 cos θ dθ
Upper limit when x = 3 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = -3 is θ = \(-\frac{\pi}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q3

Question 4.
\(\int_0^5 x^3\left(25-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Let x = 5 sin θ then
Upper limit when x = 5 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4.1

Question 5.
\(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x\)
Solution:
Take f(x) = sin8x cos7x
then f(-x) = sin8(-x) cos7(-x)
= sin8x cos7x
= f(x)
f is an even function of x.
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=2 \int_0^\pi \sin ^8 x \cos ^7 x d x\)
Now f(x) = sin8x cos7x
and f(π – x) = sin8(π – x) cos7(π – x)
= -sin8x cos7x
= -f(x)
Hence \(\int_0^\pi \sin ^8 x \cos ^7 x d x=0\)
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=0\)
[By the result that f = [0, 2a] → R is integrable on [0, a] and if f(2a – x) = -f(x) ∀ x ∈ [a, 2a] then \(\int_0^{2 a} f(x) d x=0\)]

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 6.
\(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)
Solution:
Let x = 3 cos2θ + 7 sin2θ then
dx = -6 cos θ sin θ + 14 sin θ cos θ = 8 cos θ sin θ
Upper limit when x = 7 is
7 = 3 cos2θ + 7 sin2θ
⇒ 7(1 – sin2θ) = 3 cos2θ
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
The lower limit when x = 3 is
3 = 3 cos2θ + 7 sin2θ
⇒ 3 sin2θ = 7 sin2θ
⇒ 4 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6.1

Question 7.
\(\int_2^6 \sqrt{(6-x)(x-2)} d x\)
Solution:
Put x = 2 cos2θ + 6 sin2θ
then dx = (-4 cos θ sin θ + 12 sin θ cos θ) dθ = 8 sin θ cos θ dθ
Upper limit when x = 6 is 6 = 2 cos2θ + 6 sin2θ
⇒ 6 cos2θ = 2 cos2θ
⇒ 4 cos2θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 2 is 2 = 2 cos2θ + 6 sin2θ
⇒ 2 sin2θ = 6 sin2θ
⇒ 4 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7.1

Question 8.
\(\int_0^{\frac{\pi}{2}} \tan ^5 x \cos ^8 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q8

III. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^{7 / 2}(1-x)^{5 / 2} d x\)
Solution:
Let x = sin2θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 2.
\(\int_0^\pi(1+\cos x)^3 d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2.1

Question 3.
\(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)
Solution:
Take x = 4 cos2θ + 9 sin2θ then
dx = (-8 cos θ sin θ + 18 sin θ cos θ) dθ = 10 cos θ sin θ
Upper limit when x = 9 is 9 = 4 cos2θ + 9 sin2θ
⇒ 9(1 – sin2θ) = 4 cos2θ
⇒ 5 cos2θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 4 is 4 = 4 cos2θ + 9 sin2θ
⇒ 4(1 – cos2θ) = 9 sin2θ
⇒ 5 sin2θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q3

Question 4.
\(\int_0^5 x^2(\sqrt{5-x})^7 d x\)
Solution:
Let x = 5 sin2θ then dx = 10 sin θ cos θ dθ
Upper limit when x = 5 is sin2θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin2θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

Question 5.
\(\int_0^{2 \pi}(1+\cos x)^5(1-\cos x)^3 d \dot{x}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.2

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

I.
Question 1.
Expand the following using binomial theorem.
i) (4x + 5y)5
ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
iv) (3x + x – x2)4
Solution:
i) we know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)
∴ (4x + 5y)7 = \({ }^7 \mathrm{C}_0\) (4x)7 + \({ }^7 \mathrm{C}_1\) (4x)6 (5y) + \({ }^7 \mathrm{C}_2\) (4x)5 (5y)2 + \({ }^7 \mathrm{C}_3\) (4x)4 (5y)3 + \({ }^7 \mathrm{C}_4\) (4x)3 (5y)4 + \({ }^7 \mathrm{C}_5\) (4x)2 (5y)5 + \({ }^7 \mathrm{C}_6\) (4x) (5y)6 + \({ }^7 \mathrm{C}_7\) (5y)7
= \(\sum_{r=0}^7{ }^7 C_r\) . (4x)7-r . (5y)r.

ii) We know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)

\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\) = \({ }^5 C_0\left(\frac{2}{3} x\right)^5+{ }^5 C_1\left(\frac{2}{3} x\right)^4\left(\frac{7}{4} y\right)+{ }^5 C_2\left(\frac{2}{3} x\right)^3\left(\frac{7}{4} y\right)^2 +{ }^5 C_3\left(\frac{2}{3} x\right)^3\left(\frac{7}{4} y\right)^3\) + \({ }^5 C_4\left(\frac{2}{3} x\right)\left(\frac{7}{4} y\right)^4+{ }^5 C_5\left(\frac{7}{4} y\right)^5\)

= \(\sum_{r=0}^5{ }^5 C_r \cdot\left(\frac{2}{3} x\right)^{5-r} \cdot\left(\frac{7}{4} y\right)^r\)

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) We know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 1

iv) We know that
(3 + x – x2)4 = [3 + x (1 – x)]4
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)
∴ [3 + x (1 – x)]4 = \({ }^4 C_0\) 34 + \({ }^4 C_1\)33x (1 – x + \({ }^4 C_2\) 32x2 (1 – x)2 + \({ }^4 C_3\) 3x3 (1 – x)3 + \({ }^4 C_4\) x4 (1 – x)4
= 81 + 108x (1 – x) + 54x2 (1 – 2x + x2) + 12x3 (1 – 3x + 3x2 – x3) + x4 (1 – 4x + 6x2 – 4x3 + x5
= 81 + 108x – 54x2 – 96x3 + 19x4 + 32x5 – 6x6 – 4x7 + x8.

Question 2.
Write down and simplify
i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^y\)
ii) 7th term in (3x – 4y)10
iii) 10th term \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
iv) rth term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 9).
Solution:
i) r + 1th term n the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
∴ 6th term in the expansion of \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is given by
T6 = \({ }^9 C_5 \cdot\left(\frac{2 x}{3}\right)^4 \cdot\left(\frac{3 y}{2}\right)^5\)
= 189 . x4 . y5

ii) (r + 1)th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
7th term In the expansion of (3x – 4y)10 is given by
T7 = \({ }^{10} \mathrm{C}_6\) (3x)4 (- 4y)6
= 280 . 125 x4 y6.

iii) (r + 1)th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
10th term in the expansion of \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is given by
T10 = \({ }^{14} \mathrm{C}_9\left(\frac{3 p}{4}\right)^5\) (- 5q)9
= \(\frac{-(2002) \cdot 3^5 \cdot 5^9}{4^5}\) . p5 . q9.

iv) (r + )th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
∴ rth term in the expansion of \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is given by
Tr = \({ }^8 C_{r-1}\left(\frac{3 a}{5}\right)^{9-r}\left(\frac{5 b}{7}\right)^{r-1}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
Find the number of terms in the expansion of
i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
ii) (3p + 4q)14
iii) (2x + 3y + z)7
Solution:
1) The expansion of (x + a)n contains (n + 1) terms.
∴ Expansion of \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) contains 10 terms.

ii) The expansion of (x + a)n contains (n + 1) terms.
∴ Expansion of (3p + 4q)14 contains 15 terms.

iii) Number of terms in the expansion of
(a + b + c)n = \(\frac{(n+1)(n+2)}{2}\)
∴ Expansion oF (2x + + z)7 contains \(\frac{(7+1)(7+2)}{2}\) = 36 terms.

Question 4.
Find the numerically greatest term(s) in the expansion of coefficients in (4x – 7y)49 + (4x + 7y)49.
Solution:
We know that
(4x – 7y)49 = \({ }^{49} \mathrm{C}_0\) (4x)49 – \({ }^{49} \mathrm{C}_1\)(4x)48 (7y) + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 – \({ }^{49} \mathrm{C}_3\) (4x)46 (7y)3 + ……………. + \({ }^{49} \mathrm{C}_49\) (7y)49 ………….(1)

(4x + 7y)49 = \({ }^{49} \mathrm{C}_0\) (4x)49 + \({ }^{49} \mathrm{C}_1\)(4x)48 (7y) + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 + \({ }^{49} \mathrm{C}_3\) (4x)46 (7y)3 + ……………. + \({ }^{49} \mathrm{C}_49\) (7y)49 ………….(2)

(1) + (2)
(4x – 7y)49 + (4x + 7y)49 = 2[\({ }^{49} \mathrm{C}_0\) (4x)49 + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 + \({ }^{49} \mathrm{C}_4\) (4x)45 (7y)4 + ………….. + \({ }^{49} \mathrm{C}_48\) (4x) (7y)48]
which contains 25 non-zero coefficients.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 5.
Find the sum of last 20 coefficients in the expansion of (1 + x)39.
Solution:
The last 20 coeffIcients in the expansion of (1 + x)39 are \({ }^{39} \mathrm{C}_{20},{ }^{39} \mathrm{C}_{21}, \ldots \ldots \ldots,{ }^{39} \mathrm{C}_{39}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 2

∴ The sum of last 20 coefficients in expansion of (1 + x)39 is 238.

Question 6.
If A and B are coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n – 1 respectively, then find the value of \(\frac{A}{B}\).
Solution:
Coefficient of xn in the expansion of (1 + x)2n is \({ }^{2 n} C_n\).
Coefficient of xn in the expansion of (1 + x)2n – 1 is \({ }^{2 n – 1} C_n\)
∴ A = \({ }^{2 n} C_n\) and B = \({ }^{2 n – 1} C_n\).
∴ \(\frac{A}{B}=\frac{{ }^{2 n} C_n}{2 n-1}=\frac{\frac{2 n !}{n ! n !}}{\frac{(2 n-1) !}{(n-1) ! \cdot n !}}\)
\(\frac{2 n !}{(2 n-1) ! n !} \cdot(n-1) !=\frac{2 n}{n}\)
\(\frac{A}{B}\) = 2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

II.
Question 1.
Find the coefficient of
i) x-6 in \(\left(3 x-\frac{4}{x}\right)^{10}\)
ii) x11 in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
iii) x2 in \(\left(7 x^3-\frac{2}{x^2}\right)^9\)
iv) x-7 in \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\)
Solution:
i) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
(r + 1)th term in the expansion of \(\left(3 x-\frac{4}{x}\right)^{10}\) is
Tr+1 = \({ }^{10} \mathrm{C}_{\mathrm{r}}\) . (3x)10-r . (\(\frac{-4}{x}\))r
Tr+1 = \({ }^{10} \mathrm{C}_{\mathrm{r}}\) . (3)10-r . (- 4)r . x10-2r.
for coefficient of x-6, 10 – 2r = – 6
⇒ 2r = 16
⇒ r = 8.
∴ Coefficient of x-6 is \({ }^{10} \mathrm{C}_8\) . 310-8 . (- 4)8 = 405 × 48.

ii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
(r + 1)th term in the expansion of \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is
Tr+1 = \({ }^{13} C_r \cdot\left(2 x^2\right)^{13-r} \cdot\left(\frac{3}{x^3}\right)^r\)
Tr+1 = \({ }^{13} C_r \) . 213-r . 3r . x26-5r
For coefficients of x11, 26 – 5r = 11
⇒ 5r = 15
⇒ r = 3.
Coefficient of x11 in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is \({ }^{13} C_3 \) . 213-3 . 33
= 286 . 210 . 33.

iii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
∴ General term in the expansion of \(\left(7 x^3-\frac{2}{x^2}\right)^9\) is
Tr+1 = \({ }^9 C_r \cdot\left(7 x^3\right)^{9-r}\left(\frac{-2}{x^2}\right)^r\)
= \({ }^9 C_r\) 79-r (- 2)r . x27-5r
for coefficient of x2, 27 – 5r = 2
⇒ 5r = 25
⇒ r = 5.
∴ Coefficient of x2 is \({ }^9 \mathrm{C}_5\) . 74 . (- 2)5
= – 126 . 74 . 25.

iv) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)\) is
Tr+1 = \(={ }^7 C_r \cdot\left(\frac{2}{3} x^2\right)^{7-r} \cdot\left(\frac{-5}{4 x^5}\right)^r\)
= \({ }^7 \mathrm{C}_{\mathrm{r}} \cdot\left(\frac{2}{3}\right)^{7-\mathrm{r}}\left(\frac{-5}{4}\right)^{\mathrm{r}} \mathrm{x}^{14-7 \mathrm{r}}\)
For coefficient of x-7, 14 – 7r = – 7
⇒ 7r = 21
⇒ r = 3
∴ Coefficient of x-7 is \({ }^7 C_3 \cdot\left(\frac{2}{3}\right)^4 \cdot \frac{(-5)^3}{4^3}\)
= \(-\frac{35 \times 16 \times 125}{81 \times 64}=\frac{-4375}{324}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 2.
Find the term independent of x in the expansion of
i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)
ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)
iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)
iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)
Solution:
i) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\) is
Tr+1 = \({ }^{10} C_r \cdot\left(\frac{\sqrt{x}}{3}\right)^{10-r} \cdot\left(\frac{-4}{x^2}\right)^r\)
= \({ }^{10} C_r \cdot \frac{x^{\frac{10-r}{2}}}{3^{10-r}} \cdot \frac{(-4)^r}{x^{2 r}}\)
= \({ }^{10} C_r \cdot \frac{(-4)^r}{3^{10-r}} \cdot x^{\frac{10-5 r}{2}}\)
For the independent term (i.e., the coefficient of x0)
put \(\frac{10-5 r}{2}\) = 0
⇒ r = 2.
∴ Term independent of ‘x’ in the expansion is
T3 = \({ }^{10} C_2 \frac{(-4)^2}{3^8} \times x^0=\frac{80}{729}\).

ii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\) is
Tr+1 = \({ }^{25} C_r\left(\frac{3}{\sqrt[3]{x}}\right)^{25-r} \cdot(5 \sqrt{x})^r\)
= \({ }^{25} C_r\) . 325-r . 5r . x(5r – 50)/6
For the independent term,
(i.e., coefficient of x0)
Put \(\frac{5 r-50}{6}\) = 0
⇒ r = 10.
∴ Term independent of ‘x’ in the expansion is T11 = \({ }^{25} \mathrm{C}_{10}\) . 315 . 510.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is
Tr+1 = \({ }^{14} C_r \cdot\left(4 x^3\right)^{14-r} \cdot\left(\frac{7}{x^2}\right)^r\)
= \({ }^{14} C_r\) . 414-r . 7r . x42-5r.
For the independent term, (i.e., coefficient of x0)
Put 42 – 5r = 0
which is impossible as ‘r’ is integer.
∴ Term independent of ‘x in the expansion is ‘0’.

iv) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\) is
Tr+1 = \({ }^9 C_r\left(\frac{2 x^2}{5}\right)^{9-r} \cdot\left(\frac{15}{4 x}\right)^r\)
= \({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{2}{5}\right)^{9-\mathrm{r}} \cdot\left(\frac{15}{4}\right)^{\mathrm{r}} \cdot \mathrm{x}^{18-3 \mathrm{r}}\)
For the independent term, (i.e. coefficient of x0)
Put 18 – 3r = 0
⇒ r = 6
Term independent of ‘x’ in the expansion is
T7 = \({ }^9 \mathrm{C}_6 \cdot\left(\frac{2}{5}\right)^3 \cdot\left(\frac{15}{4}\right)^6\)
= \(\frac{3^7 \cdot 5^3 \cdot 7}{2^7}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
Find the middle term(s) in the expansion of
i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)
ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)
iii) (4x2 + 5x3)17
iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Solution:
i) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3 x}{7}-2 y\right)^{10}\) is
Tr+1 = \({ }^{10} C_{\mathrm{r}} \cdot\left(\frac{3 \mathrm{x}}{7}\right)^{10-\mathrm{r}}(-2 \mathrm{y})^{\mathrm{r}}\)
The expansion has 11 (odd number) terms.
Hence T6 is the only middle term.
Thus r = 5
∴ Middle term in the expansion is
T6 = \(-10 C_5\left(\frac{6}{7}\right)^5\) . x5 . y5.

ii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion \(\left(4 a+\frac{3}{2} b\right)^{11}\) is
Tr+1 = \(\)
The expansion has 12 (even numbers) terms.
Hence T6 and T7 are the middle terms.
Thus r = 5 and r = 6
when r = 5,
T6 = \({ }^{11} C_5 \cdot(4 a)^6 \cdot\left(\frac{3}{2} b\right)^5\)
= 77 × 28 . 36 . a6 . b5.
when r = 6,
T7 = \({ }^{11} C_6(4 a)^5 \cdot\left(\frac{3}{2} b\right)^6\)
= 77 × 25 . 37 . a5 . b6.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of (4x2 + 5x3)17 is
Tr+1 = \({ }^{17} \mathrm{C}_r\) . (4x2)17-r . (5x3)r.
The expansion has 18 (even number) terms.
Hence T9 and T10 are the middle terms.
Thus r = 8 and r = 9
When r = 8.
T9 = \({ }^{17} \mathrm{C}_8\) . (4x2)9 . (5x3)8
= \({ }^{17} \mathrm{C}_8\) . 49 . 58 . x42
When r = 9,
T10 = \({ }^{17} \mathrm{C}_9\) (4x2)8 (3)9
= \({ }^{17} \mathrm{C}_9\) . 48 . 59 . x43

iv) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\) is
Tr+1 = \({ }^{20} C_r \cdot\left(\frac{3}{a^3}\right)^{20-r} \cdot\left(5 a^4\right)^r\)
The expansion has 21 (odd number) terms. Hence, T11 is the only middle term
Thus r = 10
∴ Middle term in the expansion is
T11 = \({ }^{20} C_{10} \cdot\left(\frac{3}{a^3}\right)^{10} \cdot\left(5 a^4\right)^{10}\)
= \({ }^{20} C_{10}\) . 1510 . a10.

Question 4.
Find the numerically greatest term(s) in the expansion of
i) (4 + 3x)15 when x = \(\frac{7}{2}\).
ii) (3x + 5y)12 when x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)
iii) (4a – 6b)13 when a = 3, b = 5.
iv) (3 + 7x)n when x = \(\frac{4}{5}\), n = 5.
Solution:
i) We know that (4 + 3x)15 = 415 \(\left(1+\frac{3 x}{4}\right)^{15}\)
First we find the numerically greatest term in the expansion of \(\left(1+\frac{3 x}{4}\right)^{15}\)
Let X = \(\frac{3 x}{4}\)
x = \(\frac{7}{2}\), |X| = \(\left|\frac{3}{4} \times \frac{7}{2}\right|=\frac{21}{8}\)
Now \(\frac{(n+1)|X|}{1+|X|}=\frac{16 \times \frac{21}{8}}{1+\frac{21}{8}}=\frac{336}{29}\)
not an integer.
Its integral part m = \(\left[\frac{336}{29}\right]\) = 11.
∴ Tm+1, i.e., T12 is the numerically greatest term in the binomial expansion of \(\left(1+\frac{3 x}{4}\right)^{15}\)
∴ T12 = \({ }^{15} C_{11}\left(\frac{3 x}{4}\right)^{11}\)
= \({ }^{15} \mathrm{C}_{11}\left(\frac{21}{8}\right)^{11}\)
∴ Numerically greatest term in the expansion 01(4 + 3x)15 = \({ }^{15} C_{11}\left(4^{15}\right) \frac{21^{11}}{8^{11}}\)
= \({ }^{15} C_{11} \cdot \frac{21^{11}}{2^3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

ii) We know that
(3x + 5y)12 = (3x)12 . (1 + \(\frac{5 y}{3 x}\))12
First we find the numerically greatest term in the expansion of (1 + \(\frac{5 y}{3 x}\))12

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 3

iii) We know that
(4a – 6b)13 = (4a)13 (1 – \(\frac{6 b}{4 a}\))13
First we find the numerically greatest term in the expansion of (1 – \(\frac{6 b}{4 a}\))13
Let X = – \(\frac{6 b}{4 a}\)
As a = 3 and b = 5, |X| = \(\frac{5}{2}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 4

∴ The numerically greatest terms in the expansion of (4a – 6b)13 are T10 and T11.
T10 = – 1213 . 13C9 . \(\left(\frac{5}{2}\right)^9\)
= – 13C9 . (12)4 . (30)9
and T11 = 1210 . 13C10 . \(\left(\frac{5}{2}\right)^10\)
= 143 . 217 . 313 . 510.

iv) We know that (3 + 7x)n = 3n (1 + \(\frac{7}{3}\) x)n
First we find numerically greatest term in the expansion of (1 + \(\frac{7}{3}\) x)n
Let X = \(\frac{7}{3}\) x
As x = \(\frac{4}{5}\), |X| = \(\frac{28}{15}\) and
\(\frac{(n+1)|X|}{1+|X|}=\frac{56}{5}\) and n = 15.

Its integral part,
m = \(\left[\frac{(\mathrm{n}+1)|\mathrm{X}|}{1+|\mathrm{X}|}\right]\) = 11
∴ Tm+1 i.e., T12 is numerically greatest term in the expansion of (1 + \(\frac{7}{3}\) x)n
T12 = 15C11 . \(\left(\frac{28}{15}\right)^{11}\)
∴ The numerically greatest terms in the expansion of (3 + 7x)n is
T12 = 315 . 15C11 . \(\left(\frac{28}{15}\right)^{11}\)
= 15C11 . \(\left(\frac{28}{5}\right)^{11}\) . 34.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 5.
Prove the following:
i) 2 . C0 + 5 . C1 + 8. C2 + ……………. + (3n + 2) Cn = (3n + 4) 2n-1
ii) C0 – 4 C1 + 7 . C2 – 10 . C3 + ……………. = 0,
if n is an even positive Integer.
iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)
iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3+\ldots \ldots+\frac{3^n}{n+1} C_n=\frac{4^{n+1}-1}{3(n+1)}\)
v) C0 + 2 . C1 + 4 . C2 + 8 . C3 + ………….. + 2n . Cn = 3n
Solution:
i) We know that
C0 = Cn, C1 = Cn-1, ………… Cr = Cn-r
Let S = 2 . C0 + 5 . C1 + 8 . C2 + ………….. + (3n – 1) . Cn-1 + (3n + 2) Cn ………….(1)
∴ S = (3n + 2) C0 + (3n – 1)C1 + (3n – 4)C2 + …………. + 5 . Cn-1 + 2 . Cn ………….(2)
(1) + (2) ⇒ 2S = (3n + 4) . C0 + (3n + 4) . C1 + (3n + 4) . C2 + …………… (3n + 4) . Cn
= (3n + 4) (C0 + C1 + C2 + …………… + Cn)
⇒ 2S = (3n + 4) . 2n
⇒ S = (3n + 4) . 2n-1
∴ 2 . C0 + 5 . C1 + 8 . C2 + ………… + (3n + 2) . Cn = (3n + 4) . 2n-1

ii) We know that 1, 4, 7, 10, ………… are in A.P.
(n + 1)th term,
Tn+1 = 1 + (n)3 = 3n + 1
∴ C0 – 4 . C1 + 7 . C2 – 10 . C3 + ……….. (n + 1) terms.
= \(\sum_{r=0}^n(-1)^r(3 r+1) C_r\)
= \(3 \sum_{r=0}^n(-1)^r r \cdot C_r+\sum_{r=0}^n(-1)^r \cdot C_r\)
= 3 (0) + 0 = 0
∴ C0 – 4 . C1 + 7 . C2 – 10 . C3 + ……….. = 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) Given L.H.S

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 5

iv) Given L.H.S

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 6

v) L.HS = C0 + 2 . C1 + 4 . C2 + 8 . C3 + ……….. + 2n . Cn
= C0 + C1 . 2 + C2 . 22 + C3 . 23 + ………. + Cn 2n
= (1 + 2)n
= 3n = R.H.S
∴C0 + 2 . C1 + 4 . C2 + 8 . C3 + ……….. + 2n . Cn = 3n.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 6.
Find the sum of the following:
i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \frac{{ }^{15} C_3}{{ }^{15} C_2}+\ldots+15 \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)
ii) Cn . C3 + C1 .C4 + C2 . C5 + + …………… + Cn-3 . Cn
iii) 22 C0 + 32 C1 + 42 C2 + …………. + (n + 2)2 Cn
iv) 3C0 + 6C1 + 12C2 + ……………. + 3 . 2n Cn
Solution:
i) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 7

ii) (1 + x)n = C0 + C1x + C2x2 + …………. + Cnxn …………..(1)
(x + 1)n = C0xn + C1xn – 1 + C23xn-2 + …………… + Cn …………….. (2)
(1) x (2) = (1 + x)2n
= (C0 + C1x + C2x2 + ……….. + Cnxn) (C0x + C1xn-1 + C2xn-2 + C3xn-3 + ………….. + Cn]
Comparing coefficients of xn-3 on bothsides,
2nCn-3 = C0 . C3 + C1 . C4 + C2 . C5 + …………. + Cn-3 . Cn
i.e., C0. C3 + C1 . C4 + C2. C5 + …………… + Cn-3 . Cn = 2nCn-3 = 2nCn+3
[∵ nCr = nCn-r].

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) 22 C0 + 32 C1 + 42 C2 + …………. + (n + 2)2 Cn

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 8

= n (n – 1) 2n-2 + 5 n 2n-1 + 4 (2n)
= 2n-2 [n (n – 1) + 10n + 16]
= (n2 + 9n . 16) 2n-2.

iv) 3 . C0 + 6 . C1 + 12 . C2 + ……………. + 3 . 2n Cn
= \(\sum_{r=0}^n 3 \cdot 2^r \cdot C_r\)
= \(3 \sum_{r=0}^n 2^r \cdot C_r\)
= 3 [1 + C1 (2) + C2 (22) + C3 (23) + …………. + Cn 2n]
= 3 [1 + 2]n
= 3 . 3n
= 3n+1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 7.
Using binomial theorem prove that 50n – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
We know that,
50n – 49n – 1
= (1 + 49)n – 49n – 1
= [1 + nC1 . 49 + nC2 . 492 + nC3 . 493 + ………… + 49n] – 49n – 1
= nC2 . 492 + nC3 . 493 + …………… + nCn . 49n
= 492 [nC2 + nC3 . 49 + ……….. + nCn . 49n-2]
= 492 (a positive integer)
Hence 50n 49n – 1 is divisible by 492 ∀n ∈ N.

Question 8.
Using binomial theorem prove that 54n + 52n – 1 is divisible by 676 for all positive integers n.
Solution:
We know that, 54n + 52n – 1
= (52)2n + 52n – 1
= (25)2n+ 52n – 1
= (26 – 1)2n+ 52n – 1
= 2nC0 (26)2n2nC1 (26)2n-1 + 2nC2 (26)2n-2 + …………. + 2nC2n-2(26)22nC2n-1 (26) + 2nC2n + 52n – 1
= 262n + 2nC1 (26)2n-1 + 2nC2 (26)2n-2 + …………….. + 2nC2n-2 (26)2
= 262 [262n-22nC1 262n-3 + 2nC2 262n-4 + ……… + 2nC2n-2]
= 676 (some integer)
∴ 54n + 52n – 1 is divisible by 676, ∀n ∈ N.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 9.
If (1 + x + x2)n = a0 + a1x + a2x2 + ………….. + a2nx2n, then prove that
i) a0 + a1 + a2 + ……….. + a2n = 3n
ii) a0 + a2 + a4 + ………… + a2n = \(\frac{3^n+1}{2}\)
iii) a1 + a3 + a5 + ……………. + a2n-1 = \(\frac{3^n-1}{2}\)
iv) a0 + a3 + a6 + a9 + ………… = 3n-1
Solution:
Given (1 + x + x2)n/sup> = a0 + a1x + a2x2 + ………….. + a2nx2n
i) Subtituting x = 1 in (I), we have
a0 + a1 + a2 + ……….. + a2n = 3n …………. (1)
Substituting x = – 1 in (I), we have
a0 – a1 + a2 + ……….. + a2n = 1 …………….(2)

ii) (1) + (2)
⇒ 2a0 + 2a2 + 2a4 + ……….. + 2a2n = 3n + 1
⇒ a0 + a2 + a4 + ……….. + a2n = \(\frac{3^n+1}{2}\)

iii) (1) – (2)
⇒ 2a1 + 2a3 + 2a5 + ……….. + 2a2n-1 = 3n – 1
⇒ a1 + a3 + a5 + ……….. + a2n-1 = \(\frac{3^n-1}{2}\)

iv) Substituting x = ω, in (1), we have
a0 + a1ω + a2ω2 + ……….. + a2nω2n = 0
(∵ 1 + ω + ω2 = 0)

Substituting x = ω2 in (1), we have
a0 + a1ω2 + a2ω4 + a3ω6 + ……….. + a2nω4n = 0 …………… (4)
(1) + (3) +(4):
3a0 + a1 (1 + ω + ω2) + a2 (1 + ω + ω2) + a3 (1 + ω3 + ω6) + ………… + a2n (1 + ω2n + ω4n) = 3n
⇒ 3a0 + 3a3 + 3a6 + 3a9 + …………. = 3n
(∵ 1 + ω + ω2 = 0 and ω3 = 1)
⇒ a0 + a3 + a6 + a9 + ……….. = 3n-1

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 10.
If (1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ……………+ b21 x21, then find the value of
i) b0 + b2 + b4 + ……………. + b20
ii) b1 + b3 + b5 + ……………. + b21
Solution:
Given
(1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ……………+ b21 x21 ……………(1)
Substituting x = 1 in (1),
we get b0 + b1 + b2 + ……………. + b21 = 47 ……….(2)
Substituting x = – 1 in (1),
we get b0 – b1 + b2 + ……………. – b21 = 0 …………… (3)
i) (2) + (3)
⇒ 2b0 + 2b2 + 2b4 + ……………. + b20 = 47
⇒ b0 + b2 + b4 + ……………. + b20 = 213.

ii) (2) – (3)
⇒ 2b1 + 2b3 + 2b5 + ……………. + 2b21 = 47
⇒ b1 + b3 + b5 + ……………. + b21 = 213.

Question 11.
If the coefficients of x11 and x12 in the binomial expansion of \(\left(2+\frac{8 x}{3}\right)^n\) are equal, find n.
Solution:
We know that \(\left(2+\frac{8 x}{3}\right)^n=2^n\left(1+\frac{4 x}{3}\right)^n\)
Coefficient of x11 in the expansion of \(\left(2+\frac{8 x}{3}\right)^n\) is nC11 . 2n . (\(\frac{4}{3}\))11
Coefficient of x12 in the expansion of \(\left(2+\frac{8 x}{3}\right)^n\) is nC12 . 2n . (\(\frac{4}{3}\))12
Given coefficients of x11 and x12 are same
nC11 . 2n . (\(\frac{4}{3}\))11 = nC12 . 2n . (\(\frac{4}{3}\))12
⇒ \(\frac{n !}{(n-11) ! 11 !}=\frac{n !}{(n-12) ! 12 !}\left(\frac{4}{3}\right)\)
⇒ 12 = (n – 11) \(\frac{4}{3}\)
⇒ 9 = n – 11
⇒ n = 11 + 9 = 20.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 12.
Find the remainder when 22013 is divided by 17.
Solution:
We have 22013
= 2 (22012)
= 2 (24)203
= 2 (16)503
= 2 (17 – 1)203
= 2 [503C0 17503503C1 17502 + 503C2 17501 – ……….. + 503C502 17 – 503C503]
= 2 [503C0 17503503C1 17502 + 503C2 17501 – ……….. + 503C502 17 – 503C503] – 2
= 17m – 2 where ‘m is some integer
∴ 22013 = 17m – 2 (or) 17k + 15
∴ The remainder is – 2 or 15.

Question 13.
If the coefficient of (2r + 4)th term and (3r + 4)th term in the expansion of (1 + x)21 are equal, find r.
Solution:
We know that coefficient of (r + 1)th term of (1 + x)n is nCr.
∴ Coefficient of (2r + 4)th term 0f (1 + x)21 is 21C2r+3
Also coefficient of (3r + 4)th term of (1 +x)21 is 21C3r+3.
Given coefficient of (2r + 4)th term and (3r + 4)th terms in the expansion of (1 + x)21 are equal.
21C2r+3 = 21C3r+4
∴ Either 2r + 3 = 3r + 3 0r 2r + 3 + 3 = 21
If 2r + 3 = 3r + 3 then r = 0,
If 2r + 3 + 3r = 21 then r = 3.
∴ r = 0 or r = 3.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

III.
Question 1.
If the coefficients of x9, x10, x11 in the expansion of (1 + x)n are in AP. then prove that n2 – 4m + 398 = 0.
Solution:
Coefficient of xr in the expansion of (1 + x)n is nCr.
Given coefficients of x9, x10, x11 in the expansion of (1 + x)n are in A.P., then
2(nC10) = nC9 + nC11
⇒ \(2 \frac{n !}{(n-10) ! 10 !}=\frac{n !}{(n-9) ! 9 !}+\frac{n !}{(n-11) !+11 !}\)
⇒ \(\frac{2}{10(n-10)}=\frac{1}{(n-9)(n-10)}+\frac{1}{11 \times 10}\)
⇒ \(\frac{2}{(n-10) 10}=\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\)
⇒ 22 (n – 9) = 110 + n2 – 19n + 90
⇒ n2 – 41n + 398 = 0.

Question 2.
If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + then find n.
Solution:
Let us consider nCr-1, nCr and nCr+1 as three successive binomial coefficients of (1 + xy)n.
i.e., nCr-1 = 36; nCr-1 = 84 and nCr-1 = 126
Consider \(\frac{{ }^{n_C} C_r}{{ }^n C_{r-1}}=\frac{84}{36}\)
⇒ \(\frac{n-r+1}{r}=\frac{7}{3}\)
⇒ 3n + 3 = 10r …………..(1)
Similarly,
\(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\)
⇒ \(\frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{3}{2}\)
⇒ 2n = 5r + 3 ………….(2)
⇒ 2n = 5 \(\left(\frac{3 \mathrm{n}+3}{10}\right)\) + 3 (∵ from (1))
⇒ n = 9.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)n are respectively 240, 720, 1080. find a, x, n.
Solution:
General term in the expansion of (a + x)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
∴ 2nd term,T2 = nC1 . an-1 . x …………..(1)
3rd term, T3 = nC2 . an-2 . x2 ………..(2)
4th term, T4 = nC3 . an-3 . x3 …………(3)
Given T2 = 240, T3 = 720, T3 = 1080
From (1) and (2),

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 9

Sub. n = 5 in (1), we get
5 . a4 . x = 240 ………..(6)
Substituting n = 5 in (4), we get x = \(\frac{3 a}{2}\) ………….(7)
Substituting x = \(\frac{3 a}{2}\) in (6),
we get 5 . a4 = 240
⇒ a5 = 32
⇒ a = 2
Substituting in (7), we get x = 3
∴ a = 2; x = 3 and n = 5.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 4.
If the coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are in A.P., then show that n2 – (4r + 1)n + 4r2 – 2 = 0.
Solution:
We know that, the general term, (r + 1)th term in the expansion of (1 + x)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r
∴ The coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are nCr-1, nCr and nCr+1
Given nCr-1, nCr & nCr+1 are in A.P.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 10

⇒ 2(n – r + 1) (r + 1) = r (r + 1) + (n – r) (n – r + 1)
⇒ 2nr + 2n – 2r2 – 2r + 2r + 2 = r2 + r + n2 – nr + n – nr + r2 + r
⇒ n2 – n (4r + 1) + 4r2 – 2 = 0.

Question 5.
Find the sum of the coefficients of x32 and x-18 in the expansion of (2x3 – \(\frac{3}{x^2}\))14.
Solution:
The general term in (2x3 – \(\frac{3}{x^2}\))14 is
Tr+1 = 14Cr (2x3)14-r (\(\frac{-3}{x^2}\))r
Tr+1 = (- 1)r . 14Cr . 214-r . 3r . x42-5r ………….(1)
For coefficient of x32, put
42 – 5r = 32
⇒ r = 2
Substituting r = 2 in (1), we get
T3 = 14C2 . 212 . 32 . x32
∴ Coefficient of x32 is 14C2 . 212 . 32 ………(2)
For coefficient of x-18, put
42 – 5r = 18
⇒ r=12
Substituting r = 12 in (1), we get
T13 = 14C2 . 22 . 312 . x-18
∴ Coefficient of x-18 is 14C12 . 22 . 312 ………….(3)
Hence sum of the coefficients of x32 and x-18 is
14C2 . 212 . 32 + 14C12 . 22 . 312
= 14C2 . 22 . 32 (210 + 310)
= 91 × 36(210 + 310).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 6.
If P and Q are the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)n then prove that
i) P2 – Q2 = (x2 – a2)n
ii) 4PQ (x + a)2n – (x – a)2n
Solution:
We know that,
(x + a)n = nC0 xn + nC1 xn-1 . a + nC2 xn-2 . a2 + ………….. + nCn-1 x an-1 + nCn an
= (nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………) + (nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..)
∴ (x + a)n = P + Q
∴ Sum of odd terms,
P = nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………
Sum of even terms,
Q = nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..
∴ (x + a)n = P + Q
We know that
(x – a)n = nC0 xn a – nC1 xn-1 a + nC2 xn-2 a12nC3 xn-3 a3 + ……………. + nCn (- 1)n an
= (nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………) – (nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..)
⇒ (x – a)n = P – Q

i) P2 – Q2 = (P + Q) (P – Q)
= (x + a)n (x – a)n
⇒ P2 – Q2 = (x2 – a2)n

ii) 4PQ = (P + Q)2 – (P – Q)2
= [(x + a)n]2 – [(x – a)n]2
⇒ 4PQ = (x + a)2n – (x – a)2n

Question 7.
If the coefficients of 4 consecutive terms in the expansion of (1 + x)n are a1, a2, a3, a4 respectively, then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)
Solution:
Given a1, a2, a3, a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively.
Let a1 = nCr-1
a2 = nCr
a3 = nCr+1
a4 = nCr+2

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 11

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 8.
Prove that (2nC0)2 – (2nC1)2 + (2nC2)2 – (2nC3)2 + ……….. + (2nC2n)2 = (- 1)n 2nCn.
Solution:
We know that
(x + 1)2n = 2nC0 x2n + 2nC1 x2n-1 + 2nC2 x2n-2 + …………… + 2nC2n ……………(1)
(1 – x)2n = 2nC02nC1 x + 2nC2 x2 – …………… + 2nC2n x2n ……………(2)
From (1) and (2),
(1 – x)2n (1 + x)2n = (2nC02nC1 x + 2nC2 x2 + ……………… + 2nC2n x2n)
⇒ (1 – x2)2n = [2nC0 x2n + 2nC1 x2n-1 + 2nC2 x2n-2 + …………… + 2nC2n] . [2nC02nC1 x + 2nC2 x2 + …………… + 2nC2n x2n]
Equating coefficient of x2n on both sides, we get
(- 1)n . 2nCn = (2nC0)2 + ((2nC1)2 + (2nC2)2 – (2nC3)2 + ………….. + (2nC2n)2

Question 9.
Prove that
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn) = \(\frac{(\mathbf{n}+1)^{\mathbf{n}}}{\mathbf{n} !}\) . C0 . C1 . C2 ………….. Cn.
Solution:
Given L.H.S
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 12

= R.H.S
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn) = \(\frac{(\mathbf{n}+1)^{\mathbf{n}}}{\mathbf{n} !}\) . C0 . C1 . C2 ………….. Cn.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 10.
Find the term independent of x in (1 + 3)n (1 + \(\frac{1}{3 x}\))n.
Solution:
We know that (1 + 3)n (1 + \(\frac{1}{3 x}\))n = (\(\frac{1}{3 x}\))n (1 + 3x)2n
= \(\frac{1}{3^n \cdot x^n} \sum_{r=0}^{2 n}\left({ }^{2 n} C_r\right)(3 x)^r\)
For the term, independent of x put r = n.
∴ The term independent of x in (1 + 3)n (1 + \(\frac{1}{3 x}\))n is \(\frac{1}{3^n}\) 2nCn . 3n.

Question 11.
Show that the middle term In the expansion of (1 + x)2n is \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}\) (2x)n.
Solution:
The expansion of (1 + x)2n contains 2n + 1 terms.
∴ Middle term is (n + 1)th term

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 13

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 12.
If (1 + 3x – 2x2)10 = a0 + a1x + a2x2 …………… + a20x20, then prove that
i) a0 + a1 + a2 + …………… + a20 = 210
ii) a0 – a1 + a2 – a3 + …………. + a20 = 410
Solution:
Given
(1 + 3x – 2x2)10 = a0 + a1x + a2x2 …………… + a20x20 …………….(1)
i) Put x = 1, in (1), we get
(1 + 3 – 2)10 = a0 + a1 + a2 + …………… + a20
⇒ a0 + a1 + a2 + …………… + a20 = 210.

ii) Put x – 1 in (1), we get
(1 – 3 – 2)10 = a0 – a1 + a2 – a3 + …………. + a20
a0 – a1 + a2 – a3 + …………. + a20 = 410.

Question 13.
If (3√3 + 5)2n + 1 = x and f = x – [x] (where [x] the integral part of x), find the value of x.f.
Solution:
Given (3√3 + 5)2n + 1 = x
and f = x – [x]
∴ 0 < f < 1
Let us consider F = (3√3 + 5)2n + 1
We know that
5 < 3√3 < 6 = 0 < 3√3 – 5 < 1
⇒ 0 < (3√3 – 5)2n + 1 < 1
⇒ 0 < F < 1
⇒ – 1 < – F < 0
Let x = I + f (where E is an integer)
Now
i + f – F = (3√3 + 5)2n + 1 + (3√3 – 5)2n + 1
= [2n + 1C0 (3√3)2n+1 + 2n + 1C1 (3√3)2n . 5 + 2n + 1C2 (3√3) . 52 + …………..] – [2n + 1C (3√3)2n+12n + 1C1 (3√3)2n . 5 + 2n + 1C2 (3√3)2n-1 . 52 – …………]
= 2 [2n + 1C1 (3√3) 5 + 2n + 1C3 (3√3)2n-2 53 + …………. + (2n + 1)C(2n+1) (5)2n+1]
= 2k, where k is an integer
∴ I + f – F is an even integer
⇒ f – F is also integer.
∴ (1) + (2)
⇒ – 1 < f – F < 1
⇒ f – F = 0
⇒ f = F
∴ x . f = x . F
= (3√3 + 5)2n + 1 . (3√3 – 5)2n + 1
= (27 – 25)2n + 1
= 22n + 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 14.
If R, n are posilve integers, n is odd, 0< F < 1 and if (5√5 + 11) = R + F, then prove that
i) R is an even integer and
ii) (R + F). F = 4n
Solution:
Given R, n are positive integers, 0 < F < 1 and
(5√5 + 11)n = R + F …………….(1)
i) Consider (5√5 – 11)n = f …………….(2)
We know that 11 < 5√5 < 12
⇒ 0 < 5√5 – 11 < 1
⇒0 < (5√5 – 11)n < 1
⇒ 0 < f < 1
⇒ – 1 < – f < 0 ………….(3)
From (1) and (3)
– 1 < F -f < 1 ……………(4)
Also R + F – f
= (5√5 + 11)n – (5√5 – 11)n
= [nC0 (5√5)n + nC1 (5√5)n-1 (11) + nC2 (5√5)n-2 . 112 + nC3 (5√5)n-3 . 113 + ………… + nCn (11)n] – [nC0 (5√5)nnC1 (5√5)n-1 (11) + nC2 (5√5)n-2 . 112nC3 (5√5)n-3 . 113 + ………….. + (- 1)nCn (11)n]
= 2 [nC1 (5√5)n-1 (11) + nC3 (5√5)n-3 (11)3 + ……………]
= 2k, where k is an integer (∵ f is odd)
∴ R + F – f is an even Integer.
⇒ F – f is also an interger. (∵ R is integer)
From (4), we have F – f = 0
⇒ F = f
∴ R is an even Integer.

ii) Now (R + F) . F
= (5√5 + 11)n (5√5 + 11)n
= ((5√5)2 – 112)n = 4n
∴ (R + F) . F = 4n.

Question 15.
If I, n are positive Integers, 0 < f < 1 and if (7 + 4√3) = I + f, then show that
i) I is an odd integer and
ii) (I + f) (I – f) =
Solution:
Given I and n are positve integers.
0 < f < I and (7 + 4√3)n = I + f ………….(1)

i) Let us consider (7 – 4√3)n = F
We know that 6 < 4√3 < 7
⇒ – 7 < – 4√3 < – 6
⇒ 0 < 7 – 4√3 < 1
⇒ 0 < (7 – 4√3)n < 1
⇒ 0 < F < 1
we have 0 < F + f < 2 …………(2)
Now I + f + F = (7 + 4√3)n + (7 – 4√3)n
= [nC0 7n + nC1 7n-1 (4√3) + nC2 7n-2 (4√3)2 + ………….. + nCn (4√3)n] + [nC0 7nnC1 7n-1 (4√3) + nC2 7n-2 (4√3)2 – ………….. + nCn (- 1)n (4√3)n]]
= 2[nC0 7n + nC2 7n-2 (4√3)2 + nC4 7n-4 (4√3)4 + ……………]
= 2k, where k is an integer.
∴ I + f + F is an even integer.
∴ f + F is also an integer. (∵ I is an integer)
from (2) we have f + F = I ………..(3)
∴ I + I is an even integer
⇒ I is an odd integer.

ii) Also (I + f) (I – f) = (I + f) F
= (7 + 4√3)n (7 – 4√3)n
= (72 – (4√3)2)n = 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 16.
If n is a Positive integer, prove that \(\sum_{r=1}^n \mathbf{r}^3 \cdot\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{n(n+1)^2(n+2)}{12}\).
Solution:
We know that \(\frac{{ }^{{ }^r} C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 14

Question 17.
Find the number of irrational terms in the expansion of (51/6 + 21/8)100.
Solution:
Number of terms in the expansion of (51/6 + 21/8)100 are 101.
General term in the expansion of (x + y)n is
Tr+1 = nCr xn-r . yr
∴ General term in the expansion of (51/6 + 21/8)100
Tr+1 = 100Cr . \(\left(5^{\frac{1}{6}}\right)^{100-r} \cdot\left(2^{\frac{1}{8}}\right)^r\)
= 100Cr . \(\cdot 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)
For Tr+1 to be a rational.
Clearly ‘r’ is a multiple of 8 and 100 – r is a multiple of 6
∴ r = 16, 40, 64. 88.
Number of rational terms are 4.
∴ Number of irrational terms are 101 – 4 = 97.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Exercise 7(a)

I. Evaluate the following integrals as the limit of a sum.

Question 1.
\(\int_0^5(x+1) d x\)
Solution:
We use the following formula for p = 5
and f(x) = x2 + 1, x ∈ [0, 5] and f is continuous over [0, 5].
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) I Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) I Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a)

Question 2.
\(\int_0^4 x^2 d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) I Q2

II. Evaluate the following integrals as the limit of a sum.

Question 1.
\(\int_0^4\left(x+e^{2 x}\right) d x\)
Solution:
Here p = 4, and f(x) = x + e2x
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) II Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) II Q1.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) II Q1.2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a)

Question 2.
\(\int_0^1\left(x-x^2\right) d x\)
Solution:
Here p = 1 and f(x) = x – x2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) II Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(a) II Q2.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(d)

I. Evaluate the following integrals.

Question 1.
\(\int \frac{1}{\sqrt{2 x-3 x^2+1}} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q1.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q1.2

Question 2.
\(\int \frac{\sin \theta}{\sqrt{2-\cos ^2 \theta}} d \theta\)
Solution:
Let cos θ = t, then sin θ dθ = -dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 3.
\(\int \frac{\cos x}{\sin ^2 x+4 \sin x+5} d x\)
Solution:
Let sin x = t, then cos x dx = dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q3

Question 4.
\(\int \frac{d x}{1+\cos ^2 x}\)
Solution:
Dividing by cos2x we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q4.1

Question 5.
\(\int \frac{d x}{2 \sin ^2 x+3 \cos ^2 x}\)
Solution:
Dividing by cos2x we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q5

Question 6.
\(\int \frac{1}{1+\tan x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q6
write cos x = A(sin x + cos x) + B \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x + cos x)
= A(sin x + cos x) + B(cos x – sin x)
Comparing coefficients of cos x and sin x on both sides
A – B = 0 and A + B = 1
Solving 2A = 1 ⇒ A = \(\frac{1}{2}\)
∴ B = \(\frac{1}{2}\)
∴ cos x = \(\frac{1}{2}\)(sin x + cos x) + \(\frac{1}{2}\)(cos x – sin x)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q6.1
= \(\frac{1}{2}\)x + \(\frac{1}{2}\) log|sin x + cos x| + c
[∵ sin x + cos x = t ⇒ (cos x – sin x) dx = dt in second integral]

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 7.
\(\int \frac{1}{1-\cot x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q7
Let sin x = A(sin x – cos x) + B \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x – cos x)
= A(sin x – cos x) + B(cos x + sin x)
Comparing coefficients of sin x and cos x on both sides we get
A + B = 1 and -A + B = 0
solving B = \(\frac{1}{2}\) and A = \(\frac{1}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) I Q7.1
= \(\frac{1}{2}\)x + \(\frac{1}{2}\) log|sin x – cos x| + c
[∵ sin x – cos x = t ⇒ (cos x + sin x) dx = dt in second integral]

II. Evaluate the following integrals.

Question 1.
\(\int \sqrt{1+3 x-x^2} d x\) (May ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q1.1

Question 2.
\(\int\left(\frac{9 \cos x-\sin x}{4 \sin x+5 \cos x}\right) d x\) (New Model Paper)
Solution:
Let 9 cos x – sin x = A(4 sin x + 5 cos x) + B \(\frac{\mathrm{d}}{\mathrm{dx}}\)(4 sin x + 5 cos x)
∴ 9 cos x – sin x = A(4 sin x + 5 cos x) + B(4 cos x – 5 sin x) …….(1)
Comparing coefficients of cos x and sin x on both sides
5A + 4B = 9 …….(2)
and 4A – 5B = -1 ……….(3)
Solving (2) and (3)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q2
= x + log|4 sin x + 5 cos x| + c
(∵ 4 sin x + 5 cos x = t ⇒ (4 cos x – 5 sin x) dx = dt in second integral)

Question 3.
\(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\)
Solution:
Let 2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B \(\frac{d}{d x}\)(4 cos x + 5 sin x)
= A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x) ……..(1)
Comparing coefficients of cos x and sin x on both sides we get
4A + 5B = 2 ………(2)
and 5A – 4B = 3 ………(3)
Solving (2) and (3) we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q3.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 4.
\(\int \frac{1}{1+\sin x+\cos x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q4

Question 5.
\(\int \frac{1}{3 x^2+x+1} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q5

Question 6.
\(\int \frac{d x}{\sqrt{5-2 x^2+4 x}}\)
Solution:
Consider 5 – 2x2 + 4x = 5 – (2x2 – 4x)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) II Q6.1

III. Evaluate the following integrals.

Question 1.
\(\int \frac{x+1}{\sqrt{x^2-x+1}} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q1.1

Question 2.
∫(6x + 5) \(\sqrt{6-2 x^2+x}\) dx (Mar. ’09)
Solution:
Let 6x + 5 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\)(6 – 2x2 + x) + B
= A(-4x + 1) + B
Equating the coefficients of x and constant terms
-4A = 6 ⇒ A = \(-\frac{3}{2}\)
and A + B = 5
⇒ B = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q2.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q2.2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q2.3

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 3.
\(\int \frac{d x}{4+5 \sin x}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q3.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q3.2

Question 4.
\(\int \frac{1}{2-3 \cos 2 x} d x\) (June ’10)
Solution:
Let tan x = t then sec2x dx = dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q4.1

Question 5.
∫x\(\sqrt{1+x-x^2}\) dx (May ’12)
Solution:
Let x = A \(\frac{\mathrm{d}}{\mathrm{dx}}\)(1 + x – x2) + B = A(1 – 2x) + B
Comparing the coefficient of x, constant terms on both sides
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q5
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q5.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q5.2

Question 6.
\(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\) (New Model Paper)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q6.1

Question 7.
\(\int \frac{d x}{4 \cos x+3 \sin x}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q7
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q7.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q7.2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 8.
\(\int \frac{1}{\sin x+\sqrt{3} \cos x} d x\) (May ’12)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q8.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q8.2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q8.3

Question 9.
\(\int \frac{d x}{5+4 \cos 2 x}\) (Mar. ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q9

Question 10.
\(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\) (Mar. ’11)
Solution:
Since there exist constants in both the numerator and denominator, we determine constants A, B, and C such that
2 sin x + 3 cos x + 4 = A \(\frac{\mathrm{d}}{\mathrm{dx}}\)(3 sin x + 4 cos x + 5) + B (3 sin x + 4 cos x + 5) + C
= A(3 cos x – 4 sin x) + B(3 sin x + 4 cos x + 5) + C …….(1)
Comparing both sides the coefficients of sin x, cos x, and constants
-4A + 3B = 2
⇒ 4A – 3B + 2 = 0 ……..(2)
3A + 4B – 3 = 0 ……….(3)
5B + C – 4 = 0 ………(4)
Solving (2) and (3)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q10
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q10.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q10.2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q10.3

Question 11.
\(\int \sqrt{\frac{5-x}{x-2}} d x\) on (2, 5).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q11
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q11.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q11.2

Question 12.
\(\int \sqrt{\frac{1+x}{1-x}} d x\) on (-1, 1).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q12.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 13.
\(\int \frac{d x}{(1-x) \sqrt{3-2 x-x^2}}\) on (-1, 3).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q13
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q13.1

Question 14.
\(\int \frac{d x}{(x+2) \sqrt{x+1}}\) on (-1, ∞).
Solution:
Let x + 1 = t2 then dx = 2t dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q14

Question 15.
\(\int \frac{d x}{(2 x+3) \sqrt{x+2}}\) on I ⊂ (-2, ∞) \ {\(-\frac{3}{2}\)}
Solution:
Let x + 2 = t2 then dx = 2t dt
and 2x + 3 = 2(t2 – 2) + 3 = 2t2 – 1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q15

Question 16.
\(\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x\) on (0, 1).
Solution:
Put x = t2 then dx = 2t dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q16
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q16.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q16.2

Question 17.
\(\int \frac{d x}{(x+1) \sqrt{2 x^2+3 x+1}}\) on I ⊂ R \ [-1, \(-\frac{1}{2}\)]
Solution:
Let x + 1 = \(\frac{1}{t}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q17
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q17.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 18.
\(\int \sqrt{e^x-4} d x\) on \(\left[\log _e 4, \infty\right)\).
Solution:
Let ex – 4 = t2 then ex dx = 2t dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q18
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q18.1

Question 19.
\(\int \sqrt{1+\sec x} d x\) on \(\left[\left(2 n-\frac{1}{2}\right) \pi,\left(2 n+\frac{1}{2}\right) \pi\right]\), n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q19
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q19.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)

Question 20.
\(\int \frac{d x}{1+x^4}\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q20
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q20.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d) III Q20.2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Exercise 7(b)

I. Evaluate the following definite integrals.

Question 1.
\(\int_0^a\left(a^2 x-x^3\right) d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q1.1

Question 2.
\(\int_2^3 \frac{2 x}{1+x^2} d x\) (Mar. ’12)
Solution:
Let 1 + x2 = t, then 2x dx = dt
Upper limit t = 1 + 9 = 10 when x = 3
Lower limit t = 1 + 4 = 5 when x = 2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q2
= log 10 – log 5
= log 2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 3.
\(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q3

Question 4.
\(\int_0^\pi \sin ^3 x \cos ^3 x d x\)
Solution:
I = \(\int_0^\pi \sin ^3 x \cos ^3 x d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q4

Question 5.
\(\int_0^2|1-x| d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q5

Question 6.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q6.1

Question 7.
\(\int_0^1 \frac{d x}{\sqrt{3-2 x}}\)
Solution:
Let 3 – 2x = t2 then -2 dx = 2t dt
∴ Upper limit t2 = 1 ⇒ t = 1
and Lower limit t2 = 3 ⇒ t = √3
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q7

Question 8.
\(\int_0^a(\sqrt{a}-\sqrt{x})^2 d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q8

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 9.
\(\int_0^{\frac{\pi}{4}} \sec ^4 \theta d \theta\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q9
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q9.1

Question 10.
\(\int_0^3 \frac{x}{\sqrt{x^2+16}} d x\)
Solution:
Let x2 + 16 = t2
Upper limit: x = 3
t2 = 25
⇒ t = 5
⇒ 2x dx = 2t dt
Lower limit; x = 0 ⇒ t = 4
∴ \(\int_4^5 \frac{t d t}{t}=[t]_4^5\) = 1

Question 11.
\(\int_0^1 x e^{-x^2} d x\)
Solution:
Let x2 = t then x dx = \(\frac{1}{2}\) dt
Upper limit x = 1 ⇒ t = 1
Lower limit x = 0 ⇒ t = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q11

Question 12.
\(\int_1^5 \frac{d x}{\sqrt{2 x-1}}\)
Solution:
Let 2x – 1 = t2 then 2 dx = 2t dt
⇒ dx = t dt
Upper limit when x = 5
⇒ t2 = 9
⇒ t = 3
Lower limit when x = 1
⇒ t2 = 1
⇒ t = 1 (taking positive values in [1, 5])
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) I Q12

II. Evaluate the following integrals.

Question 1.
\(\int_0^4 \frac{x^2}{1+x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q1

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 2.
\(\int_{-1}^2 \frac{x^2}{x^2+2} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q2

Question 3.
\(\int_0^1 \frac{x^2}{x^2+1} d x\) (New Model Paper, TET, Mar. ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q3

Question 4.
\(\int_0^{\frac{\pi}{2}} x^2 \sin x d x\)
Solution:
Applying integration by parts
taking u = x2 and v = sin x we get
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q4

Question 5.
\(\int_0^4|2-x| d x\) (May ’11)
Solution:
If x > 2 then |2 – x| = -(2 – x) = x – 2
If x < 2 then |2 – x| = 2 – x
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q5

Question 6.
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q6

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 7.
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x-\cos ^2 x}{\sin ^3 x+\cos ^3 x} d x\) (New Model Test Paper & Mar. ’12)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q7

Evaluate the following limits.

Question 8.
\(\lim _{n \rightarrow \infty} \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{n+n}}{n \sqrt{n}}\)
Solution:
For determining the limit we use the result that if f is continuous on [0, 1] and
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q8.1

Question 9.
\(\lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{6 n}\right]\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q9

Question 10.
\(\lim _{n \rightarrow \infty} \frac{1}{n}\left[\tan \frac{\pi}{4 n}+\tan \frac{2 \pi}{4 n}+\ldots+\tan \frac{n \pi}{4 n}\right]\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q10
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q10.1

Question 11.
\(\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{i^3}{i^4+n^4}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q11

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 12.
\(\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{i}{n^2+i^2}\)
Solution:
\(\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n^2+i^2}\)
Dividing the numerator and denominator by n2 we get
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q12
Let 1 + x2 = t then x dx = \(\frac{1}{2}\) dt
Upper limit when x = 1 is t = 2
Lower limit when x = 0 is t = 1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q12.1

Question 13.
\(\lim _{n \rightarrow \infty}\left(\frac{1+2^4+3^4+\ldots+n^4}{n^5}\right)\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q13

Question 14.
\(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \cdots \cdot\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q14
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q14.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q14.2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 15.
\(\lim _{n \rightarrow \infty}\left[\frac{(n)^{\frac{1}{n}}}{n}\right]\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) II Q15

III. Evaluate the following integrals.

Question 1.
\(\int_0^{\frac{\pi}{2}} \frac{d x}{4+5 \cos x}\) (Mar. ’93)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q1.1

Question 2.
\(\int_a^b \sqrt{(x-a)(b-x)} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q2.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q2.2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q2.3

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 3.
\(\int_0^{\frac{1}{2}} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x\)
Solution:
Let sin-1x = θ then sin θ = x and dx = cos θ dθ
Upper limit, sin θ = \(\frac{1}{2}\) ⇒ θ = \(\frac{\pi}{6}\)
Lower limit, sin θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q3.1

Question 4.
\(\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\) (Apr. ’01)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q4.1

Question 5.
\(\int_0^{\frac{\pi}{2}} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q5
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q5.1

Question 6.
\(\int_0^a x(a-x)^n d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q6

Question 7.
\(\int_0^2 x \sqrt{2-x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q7

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 8.
\(\int_0^\pi x \sin ^3 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q8.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q8.2

Question 9.
\(\int_0^\pi \frac{x}{1+\sin x} d x\) (May ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q9
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q9.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q9.2

Question 10.
\(\int_0^\pi \frac{x \sin ^3 x}{1+\cos ^2 x} d x\) (Mar. ’11)
Solution:
Let I = \(\int_0^\pi \frac{x \sin ^3 x}{1+\cos ^2 x} d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q10
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q10.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q10.2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 11.
\(\int_0^1 \frac{\log (1+x)}{1+x^2} d x\) (New Model Paper & Mar. ’10)
Solution:
Put x = tan θ then dx = sec2θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{4}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q11
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q11.1

Question 12.
\(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\) (Apr. ’99)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q12.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q12.2

Question 13.
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\cos x+\sin x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q13
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q13.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q13.2
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q13.3

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 14.
\(\int_0^\pi \frac{1}{3+2 \cos x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q14

Question 15.
\(\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q15
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q15.1

Question 16.
\(\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x\)
Solution:
We have |x sin πx| = x sin πx when -1 ≤ x ≤ 1
= -x sin πx when 1 < x ≤ \(\frac{3}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q16
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q16.1

Question 17.
\(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Solution:
Let x = tan θ then dx = sec2θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{4}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q17
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q17.1

Question 18.
\(\int_0^1 x \tan ^{-1} x d x\)
Solution:
Let x = tan θ then dx = sec2θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{4}\)
and Lower limit when x = 0 is θ = 0
∴ \(\int_0^1 x \tan ^{-1} x d x=\int_0^{\frac{\pi}{4}} \theta \tan \theta \sec ^2 \theta d \theta\)
using integration by parts by taking u = θ and v = tan θ sec2θ we get
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q18

Question 19.
\(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q19
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q19.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q19.2

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b)

Question 20.
Suppose that f : R → R is a continuous periodic function and T is its period of it. Let a ∈ R. Then prove that for any positive integer n, \(\int_a^{a+n T} f(x) d x=n \int_a^{a+T} f(x) d x\).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q20
(∵ f is a continuous function with period T)
Consider \(\int_{a+r T}^{a+(r+1) T} f(x) d x\) and (1 < r < r+1 < n)
take x = y + rT and dx = dy
Upper limit when x = a + rt + T is y = a + T
The lower limit when x = a + rT is y = a
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(b) III Q20.1
(∵ f is periodic ⇒ f(y + rT) = f(y)
Similarly, we can prove that each integral of (1) is equal to \(\int_a^{a+T} f(x) d x\)
Hence \(\int_a^{a+n T} f(x) d x=n \int_a^{a+n T} f(x) d x\)

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(c)

I. Evaluate the following integrals.

Question 1.
∫x sec2x dx on I ⊂ R – {\(\frac{(2 n+1) \pi}{2}\) : n is an integer}.
Solution:
We use the formula for integration by parts which state that
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) I Q1

Question 2.
\(\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x\), x ∈ R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) I Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 3.
\(\int \frac{\log x}{x^2} d x\) on(0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) I Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) I Q3.1

Question 4.
∫(log x)2 dx on (0, ∞).
Solution:
Take (log x)2 = u and 1 = v.
Then by integration by parts,
∫(log x)2 . 1 . dx = (log x)2 . x – ∫2 (log x) \(\frac{1}{x}\) . x dx
= x(log x)2 – 2 ∫log x . 1 . dx
= x(log x)2 – 2[logx . x – ∫\(\frac{1}{x}\) . x dx]
= x(log x)2 – 2x[log x] + 2x + c

Question 5.
∫ex (sec x + sec x tan x) dx on I ⊂ R – {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
Let f(x) = sec x then f'(x) = sec x tan x
∴ Using ∫ex [f(x) + f'(x)] dx = ex f(x) + c
we have ∫ex (sec x + sec x tan x) dx = ex sec x + c

Question 6.
∫ex cos x dx on R.
Solution:
Let I = ∫ex cos x dx
and take u = ex and v = cos x.
Then using integration by parts,
I = ex (sin x) – ∫ex sin x dx
= ex (sin x) – [ex(-cos x) – ∫ex (-cos x) dx]
= ex sin x + ex cos x – ∫ex cos x dx
= ex (sin x + cos x) – 1
2I = ex (sin x + cos x)
I = \(\frac{1}{2}\) ex (sin x + cos x) + c
∴ ∫ex cos x dx = \(\frac{1}{2}\) ex (sin x + cos x)

Question 7.
∫ex (sin x + cos x) dx on R.
Solution:
Take f(x) = sin x then f'(x) = cos x
So by using formula ∫ex [f(x) + f'(x)] dx = ex f(x) + c
we have ∫ex (sin x + cos x) dx = ex sin x + c.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 8.
∫(tan x + log sec x) ex dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π), n ∈ Z.
Solution:
Take f(x) = log|sec x| then f'(x) = \(\frac{1}{\sec x}\) (sec x tan x) = tan x
So by the formula ∫ex [f(x) + f'(x)] dx = ex f(x) + c
we have ∫(tan x + log sec x) ex dx = ex log|sec x| + c

II. Evaluate the following integrals.

Question 1.
∫xn log x dx on (0, ∞), n is a real number and n ≠ -1.
Solution:
Take u = log x and v = xn
applying integration by parts,
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q1.1

Question 2.
∫log(1 + x2) dx on R.
Solution:
Take log(1 + x2) = u and 1 = v then
using integration by parts, we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q2

Question 3.
∫√x log x dx on (0, ∞).
Solution:
Take u = log x and v = x1/2 and
using integration by parts, we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q3

Question 4.
\(\int e^{\sqrt{x}} d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q4

Question 5.
∫x2 cos x dx on R.
Solution:
Take x2 = u and cos x = v,
and using integration by parts, we get
∫x2 cos x dx = x2 (sin x) – ∫2x sin x dx
= x2 sin x – [2x(-cos x) – ∫2 . (-cos x) dx]
= x2 sin x + 2x cos x – 2∫cos x dx
= x2 sin x + 2x cos x – 2 sin x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 6.
∫x sin2x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q6

Question 7.
∫x cos2x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q7

Question 8.
∫cos√x dx on R.
Solution:
∫cos√x dx = ∫\(\frac{1}{\sqrt{x}}\) √x cos√x dx
Taking √x = t, we get \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}\) = 2 dt
∴ ∫cos √x dx = 2∫t cos t dt
using Integration by parts,
= 2[t(sin t) – ∫1 . sin t dt]
= 2[t sin t + cos t + c]
= 2[√x sin √x + cos √x ] + c

Question 9.
∫x sec22x dx on I ⊂ R \ {(2nπ + 1) \(\frac{\pi}{4}\) : n ∈ Z}
Solution:
Taking x = u and sec22x = v,
and applying integration by parts we get
∫x sec22x dx = x(\(\frac{1}{2}\) tan 2x – ∫1 . \(\frac{1}{2}\) tan 2x dx
= \(\frac{x}{2}\) tan 2x – \(\frac{1}{2}\) ∫tan 2x dx
= \(\frac{x}{2}\) tan 2x – \(\frac{1}{4}\) log|sec 2x| + c

Question 10.
∫x cot2x dx on I ⊂ R \ {nπ : n ∈ Z).
Solution:
∫x cot2x dx = ∫x(cosec2x – 1) dx = ∫x cosec2x dx – ∫x dx
Taking u = x and v = cosec2x on the first integral
and using integration by parts we get
∫x cot2x dx = x(-cot x) – ∫1 . (-cot x) dx – \(\frac{x^2}{2}\)
= -x cot x + ∫cot x dx – \(\frac{x^2}{2}\)
= -x cot x + log|sin x| – \(\frac{x^2}{2}\) + c

Question 11.
∫ex (tan x + sec2x) dx on I ⊂ R \ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
Let f(x) = tan x, then f'(x) = sec2x
So by the formula ∫ex [f(x) + f'(x)] dx = ex f(x) + c
we have ∫ex (tan x + sec2x) dx = ex tan x + c

Question 12.
∫\(e^x\left(\frac{1+x \log x}{x}\right) d x\) on (0, ∞). (Mar. ’13)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q12

Question 13.
∫eax sin bx dx on R, a, b ∈ R.
Solution:
Let I = ∫eax sin bx dx
Taking u = eax and v = sin bx
and applying integration by parts
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q13
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q13.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 14.
\(\int \frac{x e^x}{(x+1)^2} d x\) on I ⊂ R \ {-1}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q14

Question 15.
\(\int \frac{d x}{\left(x^2+a^2\right)^2}\), (a > 0) on R.
Solution:
Take substitution x = a tan θ
so that dx = a sec2θ dθ
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q15

Question 16.
∫ex log(e2x + 5ex + 6) dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q16
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q16.1

Question 17.
\(\int e^x \frac{(x+2)}{(x+3)^2} d x\) on I ⊂ R \ {-3}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) II Q17

Question 18.
∫cos(log x) dx on (0, ∞).
Solution:
Let I = ∫cos (log x) dx = ∫cos (log x) . 1 dx
Take u = cos (log x) and v = 1
and using integration by parts successively.
I = cos (log x) . x – ∫-sin (log x) \(\frac{1}{x}\) . x . dx
= x cos (log x) + ∫sin(log x) dx
= x cos (log x) + sin (log x) . x – ∫cos (log x) \(\frac{1}{x}\) . x . dx
= x cos (log x) + x . sin(log x) – ∫cos (log x) dx
= x [cos (log x) + sin (log x)] – ∫cos (log x) dx
= x [cos (log x) + sin (log x)] – I
∴ 2I = x [cos (log x) + sin (log x)]
⇒ I = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c
∴ ∫cos (log x) dx = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c

III. Evaluate the following integrals.

Question 1.
∫x tan-1x dx, x ∈ R.
Solution:
Let u = tan-1x and v = x then using integration by parts
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 2.
∫x2 tan-1x dx, x ∈ R.
Solution:
Take u = tan-1x and v = x2
and apply integration by parts we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q2.1

Question 3.
\(\int \frac{\tan ^{-1} x}{x^2} d x\), x ∈ I ⊂ R \ {0}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q3.1

Question 4.
∫x cos-1x dx, x ∈ (-1, 1).
Solution:
Let cos-1x = θ then cos θ = x
⇒ dx = -sin θ dθ
∴ ∫x cos-1x dx = ∫θ cos θ (-sin θ dθ) – \(\frac{1}{2}\) ∫θ sin 2θ dθ
Using integration by parts by taking u = θ, v = sin 2θ we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q4

Question 5.
∫x2 sin-1x dx, x ∈ (-1, 1).
Solution:
Let sin-1x = θ then sin θ = x
⇒ dx = cos θ dθ
∴ ∫x2 sin-1x dx = ∫θ sin2θ cos θ dθ
Using integration by parts
by choosing functions u = θ and v = sin2θ cos θ, we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q5

Question 6.
∫x log(1 + x) dx, x ∈ (-1, ∞).
Solution:
Take u = log(1 + x) and v = x
and apply integration by parts
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q6
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q6.1

Question 7.
∫sin√x dx on (0, ∞).
Solution:
\(\int \frac{\sqrt{x}}{\sqrt{x}} \sin \sqrt{x} d x\)
Let √x = t then \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt
∴ ∫sin √x dx = 2∫t sin t dt,
using Integration by parts by taking u = t and v = sin t, we get
= 2[t(-cos t) – ∫1 . (-cos t) dt]
= 2[-t cos t + ∫cos t dt]
= 2[-t cos t + sin t] + c
= 2[sin √x – √x cos √x ] + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 8.
∫eax sin(bx + c) dx, (a, b, c ∈ R, b ≠ 0) on R.
Solution:
Let I = ∫eax sin (bx + c) dx, taking u = eax and v = sin (bx + c)
and applying integration by parts,
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q8.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q8.2

Question 9.
∫ax cos 2x dx on R (a > 0 and a ≠ 1).
Solution:
Let I = ∫ax cos 2x dx
using integration by parts by taking cos 2x = u and ax = v, we have
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q9
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q9.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q9.2

Question 10.
\(\int \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right) d x\) on I ⊂ R – \(\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}\).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q10

Question 11.
∫sinh-1x dx on R.
Solution:
Take sinh-1x = u and v = 1,
Applying integration by parts we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q11
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q11.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)

Question 12.
∫cosh-1x dx on [1, ∞).
Solution:
∫cosh-1x dx = ∫cosh-1x . 1 . dx
Take u = cosh-1x and v = 1 then
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q12

Question 13.
∫tanh-1x dx on (-1, 1).
Solution:
∫tanh-1x dx = ∫tanh-1x . 1 . dx
Take u = tanh-1x and v = 1
and apply integration by parts we get
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c) III Q13

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(f) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(f)

I. Evaluate the following integrals.

Question 1.
∫ex (1 + x2) dx
Solution:
∫ex (1 + x2) dx = ∫(ex + ex x2) dx
= ∫ex dx + ∫ex x2 dx
= ex + x2 ex – ∫2x ex dx
= ex + x2 ex – [2x ex – ∫2ex dx]
= ex + x2 ex – 2x ex + 2ex + c
= ex [1 + x2 – 2x + 2] + c
= ex (x2 – 2x + 3) + c
[Integration by parts ∫uv dx = u∫v dx – ∫(\(\frac{\mathrm{d}}{\mathrm{dx}}\)(u) ∫v dx) is used]

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f)

Question 2.
∫x2 e-3x dx
Solution:
Use integration by parts successively using u = x2 and v = e-3x then
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) I Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) I Q2.1

Question 3.
∫x3 eax dx
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) I Q3

II.

Question 1.
Show that ∫xn e-x dx = -xn e-x + n ∫xn-1 e-x dx
Solution:
∫xn e-x dx using integration by parts
= xn (e-x) – ∫n xn-1 (-e-x) dx
= -xn e-x + n∫xn-1 e-x dx

Question 2.
If In = ∫cosnx dx, then show that In = \(\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} I_{n-2}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) II Q2

III.

Question 1.
Obtain the reduction formula for In = ∫cotnx dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot4x dx. (May ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q1.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f)

Question 2.
Obtain the reduction formula for In = ∫cosecnx dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec5x dx.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q2.1

Question 3.
If Im,n = ∫sinmx cosnx dx, then show that \(I_{m, n}=-\frac{\sin ^{m-1} x \cos ^{n+1} x}{m+n}+\frac{m-1}{m+n} I_{m-2, n}\) for a positive Integer n and an integer m ≥ 2.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q3.1
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q3.2

Question 4.
Evaluate ∫sin5x cos4x dx
Solution:
Denote I5,4 = ∫sin5x cos4x dx using the above reduction formula
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q4.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f)

Question 5.
If In = ∫(log x)n dx then show that In = x(log x)n – n In-1 and hence find ∫(log x)4 dx.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(f) III Q5
Put n = 4, 3, 2, 1 we get
I4 = x(log x)4 – 4I3
= x(log x)4 – 4[x(log x)3 – 3I2]
= x(log x)4 – 4x(log x)3 + 12[x(log x)2 – 2I1]
= x(log x)4 – 4x(log x)3 + 12x(log x)2 – 24I1]
= x(log x)4 – 4x(log x)3 + 12x(log x)2 – 24 ∫log x dx
= x(log x)4 – 4x(log x)3 + 12x(log x)2 – 24[x log x – x] + c
= x(log x)4 – 4x(log x)3 + 12x(log x)2 – 24x log x – 24x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(e) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(e)

I. Evaluate the following integrals.

Question 1.
\(\int \frac{x-1}{(x-2)(x-3)} d x\)
Solution:
\(\int \frac{x-1}{(x-2)(x-3)} d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q1
= x + 2 log|x – 3| – x – log|x – 2| + c
= 2 log|x – 3| – log|x – 2| + c
Alter: Let \(\frac{x-1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}\) (Partial fractions method)
∴ x – 1 = A(x – 3) + B(x – 2)
Comparing coefficients of x and constant terms on both sides
A + B = 1 ………(1) and
-3A – 2B = -1
⇒ 3A + 2B = 1 ………(2)
From (1), 2A + 2B = 2
Solving A = -1 and B = 2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q1.1
= -log|x – 2| + 2 log|x – 3| + c
= 2 log|x – 3| – log|x – 2| + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Question 2.
\(\int \frac{x^2}{(x+1)(x+2)^2} d x\)
Solution:
Let \(\frac{x^2}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{A}{x+2}+\frac{C}{(x+2)^2}\)
∴ x2 = A(x + 2)2 + B(x + 1)(x + 2) + C(x + 1)
Put x = -1, then A = 1
Comparing the coefficient of x2,
A + B = 1 ⇒ B = 0
Put x = -2, and we get
4 = C(-1)
⇒ C = -4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q2

Question 3.
\(\int \frac{x+3}{(x-1)\left(x^2+1\right)} d x\)
Solution:
Let \(\frac{x+3}{(x-1)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+1}\)
∴ x + 3 = A(x2 + 1) + (Bx + C)(x – 1)
Put x = 1, then 4 = 2A
⇒ A = 2
Comparing the coefficient of x2,
A + B = 0
⇒ B = -A = -2
Comparing the coefficient of constant terms
A – C = 3
⇒ C = A – 3 = 2 – 3 = -1
∴ x + 3 = 2(x2 + 1) + (-2x – 1)(x – 1) = 2(x2 + 1) – (2x + 1)(x – 1)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q3

Question 4.
\(\int \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q4

Question 5.
\(\int \frac{d x}{e^x+e^{2 x}}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q5
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q5.1

Question 6.
\(\int \frac{d x}{(x+1)(x+2)}\) (Mar. ’12; May ’11)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q6

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Question 7.
\(\int \frac{1}{e^x-1} d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q7
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q7.1

Question 8.
\(\int \frac{1}{(1-x)\left(4+x^2\right)} d x\)
Solution:
Let \(\frac{1}{(1-x)\left(4+x^2\right)}=\frac{A}{1-x}+\frac{B x+C}{x^2+4}\)
∴ 1 = A(x2 + 4) + (Bx + C)(1 – x)
When x – 1 then 5A = 1 ⇒ A = \(\frac{1}{5}\)
The coefficient of x2 on both sides gives
A – B = 0 ⇒ B = \(\frac{1}{5}\)
Comparing constant terms,
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q8

Question 9.
\(\int \frac{2 x+3}{x^3+x^2-2 x} d x\)
Solution:
x3 + x2 – 2x = x(x2 + x – 2) = x(x – 1)(x + 2)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q9
∴ 2x + 3 = A(x – 1)(x + 2) + Bx(x + 2) + Cx(x – 1)
Put x = 1 we get
5 = 3B ⇒ B = \(\frac{5}{3}\)
Put x = -2 on both sides
-4 + 3 = C(-2) (-2 – 1)
⇒ -1 = 6C
⇒ C = \(-\frac{1}{6}\)
Coefficient of x2 on both sides
A + B + C = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) I Q9.1

II. Evaluate the following integrals.

Question 1.
\(\int \frac{d x}{6 x^2-5 x+1}\)
Solution:
6x2 – 5x + 1 = 6x2 – 3x – 2x + 1
= 3x(2x – 1) – 1(2x – 1)
= (3x – 1)(2x – 1)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) II Q1

Question 2.
\(\int \frac{d x}{x(x+1)(x+2)}\)
Solution:
Let \(\frac{1}{x(x+1)(x+2)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}\)
∴ 1 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)
Put x = -1 then -B = 1 ⇒ B = -1
Coefficient of x2 both sides
A + B + C = 0 and put x = – 2 then
C(-2)(-1) = 1
⇒ C = \(\frac{1}{2}\)
∴ A = -B – C = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) II Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Question 3.
\(\int \frac{3 x-2}{(x-1)(x+2)(x-3)} d x\)
Solution:
Let \(\frac{3 x-2}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}\) + \(\frac{C}{x-3}\)
∴ 3x – 2 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
Put x = 1, then 1 = A(3)(-2) ⇒ A = \(-\frac{1}{6}\)
Put x = -2, then -8 = B(-3)(-5) ⇒ B = \(-\frac{8}{15}\)
Put x = 3, then 7 = C(2)(5) ⇒ C = \(\frac{7}{10}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) II Q3

Question 4.
\(\int \frac{7 x-4}{(x-1)^2(x+2)} d x\)
Solution:
Let \(\frac{7 x-4}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\)
∴ 7x – 4 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2
Put x = 1 both sides 3 = 3B ⇒ B = 1
Put x = -2 both sides -18 = 9C ⇒ C = -2
Comparing the coefficient of x2 on both sides
A + C = 0
⇒ A = -C = 2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) II Q4

III. Evaluate the following integrals.

Question 1.
\(\int \frac{1}{(x-a)(x-b)(x-c)} d x\)
Solution:
Let \(\frac{1}{(x-a)(x-b)(x-c)}\) = \(\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\)
Then 1 = A(x – b)(x – c) + B(x – a)(x – c) + C(x – a)(x – b)
Put x = a both sides we get 1 = A(a – b)(a – c)
⇒ A =\(\frac{1}{(a-b)(a-c)}\)
Put x = b both sides we get 1 = B(b – a)(b – c)
⇒ B = \(\frac{1}{(b-a)(b-c)}\)
Put x = c both sides we get 1 = C (c – a)(c – b)
⇒ C = \(\frac{1}{(c-a)(c-b)}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q1

Question 2.
\(\int \frac{2 x+3}{(x+3)\left(x^2+4\right)} d x\)
Solution:
Let \(\frac{2 x+3}{(x+3)\left(x^2+4\right)}=\frac{A}{x+3}+\frac{B x+C}{x^2+4}\) ……(1)
∴ 2x + 3 = A(x2 + 4) + (Bx + c)(x + 3)
Put x = -3 both sides -6 + 3 = A(9 + 4)
⇒ A = \(-\frac{3}{13}\)
Comparing the coefficient of x2 on both sides
A + B = 0
⇒ B = -A = \(\frac{3}{13}\)
and comparing constant terms on both sides
4A + 3C = 3
⇒ 3C = 3 – 4A
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q2
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q2.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Question 3.
\(\int \frac{2 x^2+x+1}{(x+3)(x-2)^2} d x\)
Solution:
Let \(\frac{2 x^2+x+1}{(x+3)(x-2)^2}=\frac{A}{x+3}+\frac{B}{x-2}\) + \(\frac{C}{(x-2)^2}\)
∴ 2x2 + x + 1 = A(x – 2)2 + B(x – 2)(x + 3) + C(x + 3)
Put x = 2 on both sides
8 + 2 + 1 = C(5)
⇒ C = \(\frac{11}{5}\)
Put x = -3 on both sides
18 – 3 + 1 = A(-5)2
⇒ A = \(\frac{16}{25}\)
Put x = 0 on both sides
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q3.1

Question 4.
\(\int \frac{d x}{x^3+1}\)
Solution:
We have x3 + 1 = (x + 1)(x2 – x + 1)
∴ \(\frac{1}{x^3+1}=\frac{1}{(x+1)\left(x^2-x+1\right)}\) = \(\frac{A}{x+1}+\frac{B x+C}{x^2-x+1}\)
∴ 1 = A(x2 – x + 1) + (Bx + C)(x + 1) …….(1)
Put x = -1 both sides 1 = A(3)
⇒ A = \(\frac{1}{3}\)
Comparing the coefficient of x2 on both sides
A + B = 0
⇒ B = \(-\frac{1}{3}\)
Comparing constant terms we get
A + C = 1
⇒ C = 1 – A
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
∴ From(1)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q4.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)

Question 5.
\(\int \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x\)
Solution:
Let cos x = t then sin x dx = -dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q5
∴ t = A(t + 1) + B(t + 2)
Put t = -1 then -1 = B(-1 + 2)
⇒ B = -1
and when t = -2 then -2 = A(-1)
⇒ A = 2
∴ \(\int \frac{t}{t^2+3 t+2} d t=\int \frac{2}{t+2} d t-\int \frac{1}{t+1} d t\)
∴ From (1)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e) III Q5.1
= log|t + 1| – 2 log|t + 2|
= log|cos x + 1| – 2 log|cos x + 2| + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e2x dx, x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q1

Question 2.
∫sin 7x dx, x ∈ R
Solution:
Let 7x = t then 7 dx = dt
⇒ dx = \(\frac{1}{7}\) dt
∴ ∫sin 7x dx = \(\frac{1}{7}\) ∫sint dt
= \(-\frac{1}{7}\) cos t
= \(-\frac{1}{7}\) cos 7x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 3.
\(\int \frac{x}{1+x^2} d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q3

Question 4.
∫2x sin(x2 + 1) dx, x ∈ R
Solution:
Let x2 + 1 = t then 2x dx = dt
∴ ∫2x sin(x2 + 1) dx = ∫sin t dt
= -cos t + c
= -cos(x2 + 1) + c

Question 5.
\(\int \frac{(\log x)^2}{x} d x\) on I ⊂ (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q5

Question 6.
\(\int \frac{e^{Tan^{-1} x}}{1+x^2} d x\) on I ⊂ (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q6

Question 7.
\(\int \frac{\sin \left({Tan}^{-1} x\right)}{1+x^2} d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q7

Question 8.
\(\int \frac{1}{8+2 x^2} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q8

Question 9.
\(\int \frac{3 x^2}{1+x^6} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q9

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 10.
\(\int \frac{2}{\sqrt{25+9 x^2}} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q10

Question 11.
\(\int \frac{3}{\sqrt{9 x^2-1}} d x\) on (\(\frac{1}{3}\), ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q11
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q11.1

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q12

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q13

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q14

Question 15.
∫sin x sin 2x sin 3x dx on R.
Solution:
Consider sin x sin 2x sin 3x = \(\frac{1}{2}\) (2 sin x sin 2x sin 3x)
= \(\frac{1}{2}\) [cos(3x – 2x) – cos(3x + 2x)] sin x
= \(\frac{1}{2}\) [sin x cos x – sin x cos 5x]
= \(\frac{1}{4}\) [2 sin x cos x – 2 sin x cos 5x]
= \(\frac{1}{4}\) [sin 2x – [sin(5x + x) + sin(x – 5x)]
= \(\frac{1}{4}\) [sin 2x – [sin 6x – sin 4x]]
= \(\frac{1}{4}\) [sin 2x – sin 6x + sin 4x]
∴ ∫sin x sin 2x sin 3x dx = \(\frac{1}{4}\) ∫sin 2x dx – \(\frac{1}{4}\) ∫sin 6x dx + \(\frac{1}{4}\) ∫sin 4x dx
= \(-\frac{1}{8}\) cos 2x + \(\frac{1}{24}\) cos 6x – \(\frac{1}{16}\) cos 4x + c
= \(\frac{1}{4}\left[\frac{\cos 6 x}{6}-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}\right]\) + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 16.
\(\int \frac{\sin x}{\sin (a+x)} d x\) on I ⊂ R – {nπ – a : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q16

II. Evaluate the following integrals.

Question 1.
\(\int(3 x-2)^{\frac{1}{2}} d x\) on (\(\frac{2}{3}\), ∞)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q1

Question 2.
\(\int \frac{1}{7 x+3} d x\) on I ⊂ R – {\(-\frac{3}{7}\)}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q2

Question 3.
\(\int \frac{\log (1+x)}{1+x} d x\) on (-1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q3

Question 4.
∫(3x2 – 4)x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q4

Question 5.
\(\int \frac{d x}{\sqrt{1+5 x}} \text { on }\left(-\frac{1}{5}, \infty\right)\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q5

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 6.
∫(1 – 2x3) x2 dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q6

Question 7.
\(\int \frac{\sec ^2 x}{(1+\tan x)^3} d x\) on I ⊂ R – {nπ – \(\frac{\pi}{4}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q7

Question 8.
∫x3 sin(x4) dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q8

Question 9.
\(\int \frac{\cos x}{(1+\sin x)^2} d x\) on I ⊂ R – {2nπ + \(\frac{3 \pi}{2}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q9

Question 10.
∫\(\sqrt[3]{\sin x}\) cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q10

Question 11.
∫2x \(e^{x^2}\) dx on R.
Solution:
Let x2 = t then 2x dx = dt
∴ ∫2x \(e^{x^2}\) dx = ∫et dt
= et + c
= \(e^{x^2}\) + c

Question 12.
\(\int \frac{e^{\log x}}{x} d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q12

Question 13.
\(\int \frac{x^2}{\sqrt{1-x^6}} d x\) on I ∈ (-1, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q13

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 14.
\(\int \frac{2 x^3}{1+x^8} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q14
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q14.1

Question 15.
\(\int \frac{x^8}{1+x^{18}} d x\) on R. (Mar. ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q15

Question 16.
\(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x\) on I ⊂ R – {x ∈ R : cos(xex) = 0}. (Mar. ’10, ’04)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q16

Question 17.
\(\int \frac{{cosec}^2 x}{(a+b \cot x)^5} d x\) on I ⊂ R – {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q17
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q17.1

Question 18.
∫ex sin ex dx on R.
Solution:
Let t = ex then dt = ex dx
∴ ∫ex sin ex dx = ∫sin t dt
= -cos t + c
= -cos(ex) + c

Question 19.
\(\int \frac{\sin (\log x)}{x} d x\) on (0, ∞).
Solution:
Let log x = t then \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{\sin (\log x)}{x} d x\) = ∫sin t dt
= -cos t + c
= -cos(log x) + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 20.
\(\int \frac{1}{x \log x} d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q20

Question 21.
\(\int \frac{(1+\log x)^n}{x} d x\) on (e-1, ∞), n ≠ -1.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q21

Question 22.
\(\int \frac{\cos (\log x)}{x} d x\) on (0, ∞).
Solution:
Let log x = t then \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{\cos (\log x)}{x} d x\) = ∫cos t dt
= sin t + c
= sin(log x) + c

Question 23.
\(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) on (0, ∞).
Solution:
Let √x = t then \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) = 2 ∫cos t dt
= 2 sin t + c
= 2 sin(√x) + c

Question 24.
\(\int \frac{2 x+1}{x^2+x+1} d x\) on R.
Solution:
Let x2 + x + 1 = t then (2x + 1) dx = dt
∴ \(\int \frac{2 x+1}{x^2+x+1} d x=\int \frac{d t}{t}\)
= log|t| + c
= log|x2 + x + 1| + c

Question 25.
\(\int \frac{a x^{n-1}}{b x^n+c} d x\) where n ∈ N, a, b, c are real numbers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xn ≠ \(-\frac{c}{b}\)}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q25

Question 26.
\(\int \frac{1}{x \log x[\log (\log x)]} d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q26

Question 27.
∫coth x dx on R.
Solution:
∫coth x dx = \(\int \frac{\cosh x}{\sinh x} d x\)
Let sinh x = t then cosh x dx = dt
∴ ∫coth x dx = \(\int \frac{\mathrm{dt}}{\mathrm{t}}\)
= log|t| + c
= log|sinh x| + c

Question 28.
\(\int \frac{1}{\sqrt{1-4 x^2}} d x \text { on }\left(-\frac{1}{2}, \frac{1}{2}\right)\).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q28

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 29.
\(\int \frac{d x}{\sqrt{25+x^2}}\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q29

Question 30.
\(\int \frac{1}{(x+3) \sqrt{x+2}}\) on I ⊂ (-2, ∞). (New Model Paper & May ’12)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q30

Question 31.
\(\int \frac{1}{1+\sin 2 x} d x\) on I ⊂ R – {\(\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}\) : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q31

Question 32.
\(\int \frac{x^2+1}{x^4+1} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q32
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q32.1

Question 33.
\(\int \frac{d x}{\cos ^2 x+\sin 2 x}\) on I ⊂ R / ({(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z} ∪ {2nπ + \({tan}^{-1} \frac{1}{2}\) : n ∈ Z})
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q33

Question 34.
\(\int \sqrt{1-\sin 2 x} d x\) on I ⊂ [2nπ – \(\frac{3 \pi}{4}\), 2nπ + \(\frac{\pi}{4}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q34
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q34.1

Question 35.
\(\int \sqrt{1+\cos 2 x} d x\) on I ⊂ [2nπ – \(\frac{\pi}{2}\), 2nπ + \(\frac{\pi}{2}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q35

Question 36.
\(\int \frac{\cos x+\sin x}{\sqrt{1+\sin 2 x}} d x\) on I ⊂ [2nπ – \(\frac{\pi}{4}\), 2nπ + \(\frac{3 \pi}{4}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q36

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 37.
\(\int \frac{\sin 2 x}{(a+b \cos x)^2} d x\) on {R, if |a| > |b|, I ⊂ {x ∈ R : a + b cos x ≠ 0}, if |a| < |b|}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q37

Question 38.
\(\int \frac{\sec x}{(\sec x+\tan x)^2} d x\) on I ⊂ R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q38
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q38.1

Question 39.
\(\int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\) on R, a ≠ 0, b ≠ 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q39

Question 40.
\(\int \frac{d x}{\sin (x-a) \sin (x-b)}\) on I ⊂ R – ({a + nπ : n ∈ Z} ∪ {b + nπ : n ∈ Z}).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q40
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q40.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 41.
\(\int \frac{1}{\cos (x-a) \cos (x-b)} \mathbf{d x}\) on I ⊂ R – ({a + \(\frac{(2 n+1) \pi}{2}\) : n ∈ Z} ∪ {b + \(\frac{(2 n+1) \pi}{2}\) : n ∈ Z})
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q41
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q41.1

III. Evaluate the following integrals.

Question 1.
\(\int \frac{\sin 2 x}{a \cos ^2 x+b \sin ^2 x} d x\) on I ⊂ R – {x ∈ R | a cos2x + b sin2x = 0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q1

Question 2.
\(\int \frac{1-\tan x}{1+\tan x} d x\) for x ∈ I ⊂ R – {nπ – \(\frac{\pi}{4}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q2

Question 3.
\(\int \frac{\cot (\log x)}{x} d x\), x ∈ I ⊂ (0, ∞) – {e : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q3

Question 4.
∫ex cot ex dx, x ∈ I ⊂ R – {log nπ : n ∈ Z}
Solution:
Let ex = t then ex dx = dt
∴ ∫ex cot x dx = ∫cot t dt
= log|sin t| + c
= log|sin(ex)| + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 5.
∫sec(tan x) sec2x dx on I ⊂ {x ∈ E : tan x ≠ \(\frac{(2 k+1) \pi}{2}\) for any k ∈ Z), where E = R – {\(\frac{(2 n+1) \pi}{2}\), n ∈ Z}
Solution:
Let tan x = t, then sec2x dx = dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q5

Question 6.
\(\int \sqrt{\sin x} \cos x d x\) on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q6

Question 7.
∫tan4x sec2x dx, x ∈ I ⊂ R – {\(\frac{(2 n+1) x}{2}\) : n ∈ Z}.
Solution:
Let tan x = t then sec2x dx = dt
∴ ∫tan4x sec2x dx = ∫t4dt
= \(\frac{t^5}{5}\) + c
= \(\frac{\tan ^5 x}{5}\) + c

Question 8.
\(\int \frac{2 x+3}{\sqrt{x^2+3 x-4}} d x\), x ∈ I ⊂ R – [-4, 1]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q8

Question 9.
\(\int {cosec}^2 x \sqrt{\cot x} d x\) on (0, \(\frac{\pi}{2}\)].
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q9

Question 10.
∫sec x log(sec x + tan x) dx on (0, \(\frac{\pi}{2}\)).
Solution:
Let log(sec x + tan x) = t then \(\frac{1}{\sec x+\tan x}\) (sec x tan x + sec2x) dx = dt
⇒ \(\frac{\sec x(\sec x+\tan x) d x}{\sec x+\tan x}\) = dt
⇒ sec x dx = dt
∴ ∫sec x log(sec x + tan x) dx = ∫t dt + c
= \(\frac{\mathrm{t}^2}{2}\) + c
= \(\frac{1}{2}\) [log(sec x + tan x)]2 + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 11.
∫sin3x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin3x
⇒ 4 sin3x = 3 sin x – sin 3x
⇒ sin3x = \(\frac{3 \sin x-\sin 3 x}{4}\)
∴ ∫sin3x dx = \(\frac{1}{2}\) ∫(3 sin x – sin 3x) dx
= \(\frac{3}{4}\)∫sin x dx – \(\frac{1}{4}\)∫sin 3x dx
= \(-\frac{3}{4}\) cos x + \(\frac{1}{12}\) cos 3x + c
= \(\frac{1}{12}\) (cos 3x – 9 cos x] + c

Question 12.
∫cos3x dx on R.
Solution:
cos 3x = 4 cos3x – 3 cos x
⇒ 4cos3x = cos 3x + 3 cos x
⇒ cos3x = \(\frac{1}{4}\)(cos 3x + 3 cos x)
∴ ∫cos3x dx = \(\frac{1}{4}\)∫(cos 3x + 3 cos x) dx
= \(\frac{1}{12}\) sin 3x + \(\frac{3}{4}\) sin x dx + c
= \(\frac{1}{12}\) [sin 3x + 9 sin x] + c

Question 13.
∫cos x cos 2x dx on R.
Solution:
We have cos A cos B = \(\frac{1}{2}\) [cos(A + B) + cos(A – B)]
⇒ cos x cos 2x = \(\frac{1}{2}\) [cos 3x + cos x]
∴ ∫cos x cos 2x dx = \(\frac{1}{2}\) [∫cos 3x + ∫cos x] dx
= \(\frac{1}{2}\left(\frac{1}{3}\right)\) sin 3x + \(\frac{1}{2}\) sin x + c
= \(\frac{1}{6}\) sin 3x + \(\frac{1}{2}\) sin x + c
= \(\frac{1}{6}\) [sin 3x + 3 sin x] + c

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos x cos 3x = \(\frac{1}{2}\) [cos 4x + cos 2x]
∴ ∫cos x cos 3x dx = \(\frac{1}{2}\) ∫cos 4x dx + \(\frac{1}{2}\) ∫cos 2x dx
= \(\frac{1}{8}\) sin 4x + \(\frac{1}{4}\) sin 2x + c

Question 15.
∫cos4x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q15

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 16.
\(\int x \sqrt{4 x+3} d x\) on (\(-\frac{3}{4}\), ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q16

Question 17.
\(\int \frac{d x}{\sqrt{a^2-(b+c x)^2}}\) on {x ∈ R : |b + cx| < a}, where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:
Let b + cx = t then c dx = dt
⇒ dx = \(\frac{1}{c}\) dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q17

Question 18.
\(\int \frac{d x}{a^2+(b+c x)^2}\) on R, where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q18

Question 19.
\(\int \frac{d x}{1+e^x}\), x ∈ R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q19

Question 20.
\(\int \frac{x^2}{(a+b x)^2} d x\), x ∈ I ⊂ R – {\(-\frac{a}{b}\)}, where a, b are real numbers, b ≠ 0.
Solution:
Let a + bx = t then b dx = dt
⇒ dx = \(\frac{1}{b}\) dt
Also x = \(\frac{t-a}{b}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q20

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

Question 21.
\(\int \frac{x^2}{\sqrt{1-x}} d x\), x ∈ (-∞, 1).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q21

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(a)

I. Evaluate the following integrals.

Question 1.
∫(x3 – 2x2 + 3) dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q1

Question 2.
∫2x√x dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q2

Question 3.
\(\int \sqrt[3]{2 x^2}\) dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q3

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 4.
\(\int\left(\frac{x^2+3 x-1}{2 x}\right)\) dx, x ∈ I ⊂ R – {0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q4.1

Question 5.
\(\int \frac{1-\sqrt{x}}{x}\) dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q5

Question 6.
\(\int\left(1+\frac{2}{x}-\frac{3}{x^2}\right) d x\) on I ⊂ R – {0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q6

Question 7.
\(\int\left(x+\frac{4}{1+x^2}\right) d x\) on R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q7

Question 8.
\(\int\left(e^x-\frac{1}{x}+\frac{2}{\sqrt{x^2-1}}\right) d x\) on I ⊂ R – [-1, 1]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q8.1

Question 9.
\(\int\left(\frac{1}{1-x^2}+\frac{1}{1+x^2}\right) d x\) on (-1, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q9

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 10.
\(\int\left(\frac{1}{\sqrt{1-x^2}}+\frac{2}{\sqrt{1+x^2}}\right) d x\) on (-1, 1). [May ’11]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q10

Question 11.
\(\int e^{\log \left(1+\tan ^2 x\right)} d x\) on I ⊂ R – {\(\frac{(2 n+1) \pi}{2}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q11

Question 12.
\(\int \frac{\sin ^2 x}{1+\cos 2 x} d x\) on I ⊂ R – {(2n ± 1)π : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q12.1

II. Evaluate the following integrals.

Question 1.
∫(1 – x2)3 dx on (-1, 1).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q1

Question 2.
\(\int\left(\frac{3}{\sqrt{x}}-\frac{2}{x}+\frac{1}{3 x^2}\right) d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 3.
\(\int\left(\frac{\sqrt{x}+1}{x}\right)^2 d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q3.1

Question 4.
\(\int \frac{(3 x+1)^2}{2 x} d x\), x ∈ I ⊂ R – {0}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q4

Question 5.
\(\int\left(\frac{2 x-1}{3 \sqrt{x}}\right)^2 d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q5

Question 6.
\(\int\left(\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x^2-1}}-\frac{3}{2 x^2}\right) d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q6

Question 7.
∫(sec2x – cos x + x2) dx, x ∈ I ⊂ R – {\(\frac{n \pi}{2}\) : n is an odd integer}.
Solution:
∫(sec2x – cos x + x2) dx
= ∫sec2x dx – ∫cos x dx + ∫x2 dx
= tan x – sin x + \(\frac{x^3}{3}\) + c

Question 8.
∫(sec x tan x + \(\frac{3}{x}\) – 4] dx, x ∈ I ⊂ R – ({\(\frac{n \pi}{2}\) : n is an odd integer} ∪ {0})
Solution:
∫(sec x tan x + \(\frac{3}{x}\) – 4] dx
= ∫sec x + tan x dx + 3 ∫\(\frac{dx}{x}\) – 4∫dx
= sec x + 3 log|x| – 4x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 9.
\(\int\left(\sqrt{x}-\frac{2}{1-x^2}\right) d x\) on (0, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q9
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q9.1

Question 10.
\(\int\left(x^3-\cos x+\frac{4}{\sqrt{x^2+1}}\right) d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q10

Question 11.
\(\int\left(\cosh x+\frac{1}{\sqrt{x^2+1}}\right) d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q11

Question 12.
\(\int\left(\sinh x+\frac{1}{\left(x^2-1\right)^{\frac{1}{2}}}\right) d x\), x ∈ I ⊂ (-∞, -1) ∪ (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q12.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 13.
\(\int \frac{\left(a^x-b^x\right)^2}{a^x b^x} d x\), (a > 0, a ≠ 1 and b > 0, b ≠ 1) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q13

Question 14.
∫sec2x cosec2x dx on I ⊂ R – ({nx : n ∈ Z} ∪ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}) (May ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q14
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q14.1

Question 15.
\(\int\left(\frac{1+\cos ^2 x}{1-\cos 2 x}\right) d x\) on I ⊂ R – {nπ : n ∈ Z} (Mar. ’13)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q15

Question 16.
\(\int \sqrt{1-\cos 2 x} d x\) on I ⊂ [2nπ, (2n + 1)π], n ∈ Z. (Mar. ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q16

Question 17.
\(\int \frac{1}{\cosh x+\sinh x} d x\) on R.
Solution:
\(\int \frac{1}{\cosh x+\sinh x} d x\)
= \(\int \frac{\cosh x-\sinh x}{\cosh ^2 x-\sinh ^2 x} d x\)
= ∫(cos hx – sin hx) dx (∵ cosh2x – sinh2x = 1)
= ∫cosh x dx – ∫sinh x dx
= sinh x – cosh x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 18.
\(\int \frac{1}{1+\cos x} d x\) on I ⊂ R – [(2n + 1)π : n ∈ Z].
Solution:
\(\int \frac{1}{1+\cos x} d x\)
= \(\int \frac{1-\cos x}{(1+\cos x)(1-\cos x)} d x\)
= \(\int \frac{1-\cos x}{\sin ^2 x} d x\)
= ∫cosec2x dx – ∫cot x cosec x dx
= -cot x + cosec x + c
= cosec x – cot x + c