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:

Question 2.
\(\int \frac{\sin \theta}{\sqrt{2-\cos ^2 \theta}} d \theta\)
Solution:
Let cos θ = t, then sin θ dθ = -dt

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

Question 4.
\(\int \frac{d x}{1+\cos ^2 x}\)
Solution:
Dividing by cos²x we get

Question 5.
\(\int \frac{d x}{2 \sin ^2 x+3 \cos ^2 x}\)
Solution:
Dividing by cos²x we get

Question 6.
\(\int \frac{1}{1+\tan x} d x\)
Solution:

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)

= \(\frac{1}{2}\)x + \(\frac{1}{2}\) log|sin x + cos x| + c
[∵ sin x + cos x = t ⇒ (cos x – sin x) dx = dt in second integral]

Question 7.
\(\int \frac{1}{1-\cot x} d x\)
Solution:

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 = \(\frac{1}{2}\)

= \(\frac{1}{2}\)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:

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)

= 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

Question 4.
\(\int \frac{1}{1+\sin x+\cos x} d x\)
Solution:

Question 5.
\(\int \frac{1}{3 x^2+x+1} d x\)
Solution:

Question 6.
\(\int \frac{d x}{\sqrt{5-2 x^2+4 x}}\)
Solution:
Consider 5 – 2x² + 4x = 5 – (2x² – 4x)

III. Evaluate the following integrals.

Question 1.
\(\int \frac{x+1}{\sqrt{x^2-x+1}} d x\)
Solution:

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 + \(\frac{3}{2}\) = \(\frac{13}{2}\)

Question 3.
\(\int \frac{d x}{4+5 \sin x}\)
Solution:

Question 4.
\(\int \frac{1}{2-3 \cos 2 x} d x\) (June ’10)
Solution:
Let tan x = t then sec²x dx = dt

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

Question 6.
\(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\) (New Model Paper)
Solution:

Question 7.
\(\int \frac{d x}{4 \cos x+3 \sin x}\)
Solution:

Question 8.
\(\int \frac{1}{\sin x+\sqrt{3} \cos x} d x\) (May ’12)
Solution:

Question 9.
\(\int \frac{d x}{5+4 \cos 2 x}\) (Mar. ’11)
Solution:

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)

Question 11.
\(\int \sqrt{\frac{5-x}{x-2}} d x\) on (2, 5).
Solution:

Question 12.
\(\int \sqrt{\frac{1+x}{1-x}} d x\) on (-1, 1).
Solution:

Question 13.
\(\int \frac{d x}{(1-x) \sqrt{3-2 x-x^2}}\) on (-1, 3).
Solution:

Question 14.
\(\int \frac{d x}{(x+2) \sqrt{x+1}}\) on (-1, ∞).
Solution:
Let x + 1 = t² then dx = 2t dt

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

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

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 = \(\frac{1}{t}\)

Question 18.
\(\int \sqrt{e^x-4} d x\) on \(\left[\log _e 4, \infty\right)\).
Solution:
Let e^x – 4 = t² then ex dx = 2t dt

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:

Question 20.
\(\int \frac{d x}{1+x^4}\) on R.
Solution: