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:

Question 2.

\(\int_0^{\frac{\pi}{2}} \cos ^{11} x d x\)

Solution:

Question 3.

\(\int_0^{\frac{\pi}{2}} \cos ^7 x \sin ^2 x d x\)

Solution:

\(\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x\)

Question 4.

\(\int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^4 x d x\)

Solution:

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)

Question 6.

\(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\)

Solution:

Take f(x) = sin^{2}x cos^{4}x

Then f(π – x) = sin^{2}(π – x) cos^{4}(π – x)

= sin^{2}x cos^{4}x

= f(x)

Question 7.

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta \cos ^7 \theta d \theta\)

Solution:

f(θ) = sin^{2}θ cos^{7}θ

f(-θ) = sin^{2}(-θ) cos^{7}(-θ)

= sin^{2}θ cos^{7}θ

= f(θ); and f is even

Question 8.

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta\)

Solution:

Let f(θ) = sin^{3}θ cos^{3}θ

∴ f(-θ) = sin^{3}(-θ) cos^{3}(-θ)

= -sin^{3}θ cos^{3}θ

= -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

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

Question 10.

\(\int_0^2 x^{\frac{3}{2}} \sqrt{2-x} d x\)

Solution:

Take x = 2 sin^{2}θ, then dx = 4 sin θ cos θ dθ

Upper limit when x = 2 is sin^{2}θ = 1 ⇒ θ = \(\frac{\pi}{2}\)

Lower limit when x = 0 is sin^{2}θ = 0 ⇒ θ = 0

II. Evaluate the following integrals.

Question 1.

\(\int_0^1 x^5(1-x)^{\frac{5}{2}} d x\)

Solution:

Let x = sin^{2}θ 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

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

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 = \(-\frac{\pi}{2}\)

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

Question 5.

\(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x\)

Solution:

Take f(x) = sin^{8}x cos^{7}x

then f(-x) = sin^{8}(-x) cos^{7}(-x)

= sin^{8}x cos^{7}x

= 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) = sin^{8}x cos^{7}x

and f(π – x) = sin^{8}(π – x) cos^{7}(π – x)

= -sin^{8}x cos^{7}x

= -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 \(\int_0^{2 a} f(x) d x=0\)]

Question 6.

\(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)

Solution:

Let x = 3 cos^{2}θ + 7 sin^{2}θ 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 cos^{2}θ + 7 sin^{2}θ

⇒ 7(1 – sin^{2}θ) = 3 cos^{2}θ

⇒ cos θ = 0

⇒ θ = \(\frac{\pi}{2}\)

The lower limit when x = 3 is

3 = 3 cos^{2}θ + 7 sin^{2}θ

⇒ 3 sin^{2}θ = 7 sin^{2}θ

⇒ 4 sin^{2}θ = 0

⇒ sin θ = 0

⇒ θ = 0

Question 7.

\(\int_2^6 \sqrt{(6-x)(x-2)} d x\)

Solution:

Put x = 2 cos^{2}θ + 6 sin^{2}θ

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 cos^{2}θ + 6 sin^{2}θ

⇒ 6 cos^{2}θ = 2 cos^{2}θ

⇒ 4 cos^{2}θ = 0

⇒ cos θ = 0

⇒ θ = \(\frac{\pi}{2}\)

Lower limit when x = 2 is 2 = 2 cos^{2}θ + 6 sin^{2}θ

⇒ 2 sin^{2}θ = 6 sin^{2}θ

⇒ 4 sin^{2}θ = 0

⇒ sin θ = 0

⇒ θ = 0

Question 8.

\(\int_0^{\frac{\pi}{2}} \tan ^5 x \cos ^8 x d x\)

Solution:

III. Evaluate the following integrals.

Question 1.

\(\int_0^1 x^{7 / 2}(1-x)^{5 / 2} d x\)

Solution:

Let x = sin^{2}θ then dx = 2 sin θ cos θ dθ

Upper limit when x = 1 is sin^{2}θ = 1 ⇒ θ = \(\frac{\pi}{2}\)

Lower limit when x = 0 is sin^{2}θ = 0 ⇒ θ = 0

Question 2.

\(\int_0^\pi(1+\cos x)^3 d x\)

Solution:

Question 3.

\(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)

Solution:

Take x = 4 cos^{2}θ + 9 sin^{2}θ 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 cos^{2}θ + 9 sin^{2}θ

⇒ 9(1 – sin^{2}θ) = 4 cos^{2}θ

⇒ 5 cos^{2}θ = 0

⇒ cos θ = 0

⇒ θ = \(\frac{\pi}{2}\)

Lower limit when x = 4 is 4 = 4 cos^{2}θ + 9 sin^{2}θ

⇒ 4(1 – cos^{2}θ) = 9 sin^{2}θ

⇒ 5 sin^{2}θ = 0

⇒ sin θ = 0

⇒ θ = 0

Question 4.

\(\int_0^5 x^2(\sqrt{5-x})^7 d x\)

Solution:

Let x = 5 sin^{2}θ then dx = 10 sin θ cos θ dθ

Upper limit when x = 5 is sin^{2}θ = 1 ⇒ θ = \(\frac{\pi}{2}\)

Lower limit when x = 0 is sin^{2}θ = 0 ⇒ θ = 0

Question 5.

\(\int_0^{2 \pi}(1+\cos x)^5(1-\cos x)^3 d \dot{x}\)

Solution: