Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c)

I. Evaluate the following definite integrals.

Question 1.

\(\int_{\pi/2}^{\pi/2}\)sin^{10} x dx

Solution:

Question 2.

\(\int_0^{\pi/2}\)cos^{11} x dx

Solution:

Question 3.

\(\int_0^{\pi/2}\)cos^{7} x . sin²x dx.

Solution:

Question 4.

\(\int_0^{\pi/2}\)sin^{4} x . cos^{4} x dx.

Solution:

Question 5.

\(\int_0^{\pi/2}\)sin³ x cos^{6} x dx.

Solution:

\(\int_0^{\pi/2}\)sin³ x cos^{6} x dx.

\(\int_0^{\pi/2}\)(1 – cos² x) cos^{6} x.sin x dx

Question 6.

\(\int_0^{2\pi}\)sin² x cos^{4} x dx.

Solution:

Question 7.

\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos^{7} θ dθ.

Solution:

sin² θ cos^{7} θ is even function

f(θ) = sin² θ . cos^{7} θ dθ

f(-θ) = sin² (-θ) . cos^{7} (-θ)

= f(θ)

= 2\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos^{7} θ dθ

\(\int_0^{\pi/2}\)sin^{m} x cos^{n}x dx

n is odd n = 7

Question 8.

\(\int_{-\pi/2}^{\pi/2}\)sin³ θ .cos³ θ dθ.

Solution:

f(θ) = sin³ θ . cos³ θ dθ

f(-θ) = sin³ (-θ) . cos³ (-θ)

= -sin³ θ cos³ θ = -f(θ)

f(θ) is odd

∴ \(\int_{-\pi/2}^{\pi/2}\)sin³ θ.cos³ θ dθ = 0

Question 9.

\(\int_0^a\)x(a² – x²)^{7/2} dx

Solution:

x = a sin θ, a = a sin θ

dx = a cos θ dθ, θ = π/2

= \(\int_0^{\pi/2}\)a sin θ(a² – a²sin²θ)^{7/2} a cos θ dθ

= \(\int_0^{\pi/2}\)a^{9} cos^{8} θ sin θ dθ

= a^{9}\(\int_0^{\pi/2}\)cos^{8} . θ sin θ dθ

Question 10.

\(\int_0^2\)x^{3/2}.\(\sqrt{2-x}\)dx

Solution:

x = 2 cos² θ

dx = 4 cos θ sin θ dθ

II. Evaluate the following integrals.

Question 1.

\(\int_0^1\)x^{5}(1 – x)^{3/2} dx

Solution:

x = sin² θ

dx = 2 sin θ . cos θ . dθ

Question 2.

\(\int_0^4\)(16 – x²)^{5/2} dx

Solution:

Question 3.

\(\int_{-3}^3\)(9 – x²)^{3/2} x dx

Solution:

Let f(x) = (9 – x²)^{3/2}_{x}

f(x) = (9 – (-x²))^{3/2}(-x)

= (9 – x²)^{3/2} . x

= -f(x)

∴ f is odd function

∴ \(\int_{-3}^3\)(9 – x²)^{3/2} x dx = 0

Question 4.

\(\int_0^5\)x³(25 + x²)^{7/2} dx

Solution:

Let I = \(\int_0^5\)x³(25 + x²)^{7/2} dx

Put x = 5 sin θ

dx = 5 cosθ dθ

Question 5.

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

Solution:

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

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

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

∴ f is even function.

∴ \(\int_{-\pi}^{\pi}\)sin^{8} x cos^{7} x dx = 2\(\int_0^{\pi}\)sin^{8} x cos^{7} x = 0

Question 6.

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

Solution:

Put x = 3 cos²θ + 7 sin²θ

dx = (7 – 3)sin2θ dθ

dx = 4 sin 2θ dθ

U.L.

x = 3 cos²θ + 7 sin²θ

7 = 3 cos²θ + 7 sin²θ

4 cos²θ = 0

θ = \(\frac{\pi}{2}\)

L.L

x = 3 cos²θ + 7 sin²θ

3 = 3 sin²θ + 7 sin²θ

4 sin²θ = 0

sinθ = 0

θ = 0

7 – x = 7 – (3 cos²θ + 7 sin²θ)

= (7 – 3)cos²θ

= 4 cos²θ

x – 3 = 3 cos²θ + 7 sin²θ – 3

= (7 – 3)sin²θ

= 4 sin²θ

Question 7.

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

Solution:

Put x = 2 cos²θ + 6 sin²θ

dx = (6 – 2) sin2θ dθ

dx = 4 sin2θ dθ

U.L

x = 2 cos²θ + 6 sin²θ

6 = 2 cos²θ + 6 sin²θ

4 cos²θ = 0

cos θ = 0

θ = \(\frac{\pi}{2}\)

L.L

x = 2 cos²θ + 6 sin²θ

2 = 2 cos²θ + 6 sin²θ

4 sin²θ = 0

θ = 0

6 – x = 6 – (2 cos²θ + 6 sin²θ)

= (6 – 2) cos²θ

= 4 cos²θ

x – 2 = 2 cos²θ + 6 sin²θ – 2

= (6 – 2)sin²θ

= 4 sin²θ

Question 8.

\(\int_0^{\pi}\)tan^{5}x cos^{8}x dx

Solution:

III. Evaluate the following integrals.

Question 1.

\(\int_0^1\)x^{7/2} (1 – x)^{5/2} dx

Solution:

Put x = sin²θ

dx = 2 sin θ cos θ dθ

U.L

x = sin²θ

1 = sin²θ

θ = \(\frac{\pi}{2}\)

L.L

x = sin²θ

0 = sin²θ

θ = 0

Question 2.

\(\int_0^{\pi}\)(1 + cos x)³ dx

Solution:

Question 3.

\(\int_4^9\frac{dx}{\sqrt{(9 – x)(x – 4)}}\)

Solution:

Put x = 4 cos²θ + 9 sin²θ

dx = (9 – 4)sin2θ dθ

dx = 5 sin2θ dθ

U.L

x = 4 cos²θ + 9 sin²θ

9 = 4 cos²θ + 9 sin²θ

5 cos²θ = 0

θ = \(\frac{\pi}{2}\)

L.L

x = 4 cos²θ + 9 sin²θ

4 = 4 cos²θ + 9 sin²θ

5 sin²θ = 0

θ = 0

9 – x = 9 – (4 cos²θ + 9 sin²θ)

= (9 – 4) cos²θ

= 5 cos²θ

x – 4 = 4 cos²θ + 9 sin²θ – 4

= (9 – 4) sin²θ

= 5 sin²θ

Question 4.

\(\int_0^5\)x²(\(\sqrt{5-x}^7\) dx

Solution:

Put x = 5 sin²θ

dx = 10 sinθ cosθ dθ

U.L

x = 5 sin²θ

5 = 5 sin²θ

sin²θ = 1

θ = \(\frac{\pi}{2}\)

L.L

x = 5 sin²θ

0 = 5sin²θ

sin²θ = 0

θ = 0

Question 5.

\(\int_0^{2\pi}\)(1 + cos x)^{5}(1 – cos x)³ dx.

Solution:

\(\int_0^{2\pi}\)(1 + cos x)^{5}(1 – cos x)³ dx . (1 + cos x)³(1 + cos x)²(1 – cos x)³