Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 7 Integrals Ex 7.10 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.10

Evaluate the following integrals:

Question 1.

\(\int_{0}^{1}\) \(\frac{x}{x^{2}+1}\) dx

Solution:

Let I = \(\int_{0}^{1}\) \(\frac{x}{x^{2}+1}\) dx.

Put x^{2} + 1 = t, 2x dx = dt.

When x = 1, t = 2 and when x = 0, t = 1.

∴ I = \(\int_{2}^{1}\) \(\frac{dt}{t}\)

= log 2.

Question 2.

\(\int_{0}^{\frac{\pi}{2}}\) \(\sqrt{sinϕ}\)cos^{5}ϕdϕ

Solution:

Let I = \(\int_{0}^{\frac{\pi}{2}}\) \(\sqrt{sinϕ}\)cos^{5}ϕdϕ

= \(\int_{0}^{\frac{\pi}{2}}\) \(\sqrt{sinϕ}\)cos^{4}ϕcosϕdϕ

Put sinϕ = t so that cosϕ dϕ = dt.

When ϕ = 0, t = 0 and when ϕ = \(\frac{π}{2}\), t = 1.

Question 3.

\(\int_{0}^{1}\) sin^{-1}(\(\frac{2 x}{1+x^{2}}\))dx

Solution:

Integrating by parts, taking θ as first function,

Question 4.

\(\int_{0}^{2}\)x\(\sqrt{x+2}\) dx

Solution:

Let I = \(\int_{0}^{2}\)x\(\sqrt{x+2}\) dx

Put x + 2 = t^{2} so that dx = 2t dt.

When x = 0, t = \(\sqrt{2}\) and when x = 2, t^{2} = 4 ⇒ t = 2.

Question 5.

\(\int_{0}^{\frac{\pi}{2}}\) \(\frac{\sin x}{1+\cos ^{2} x}\) dx

Solution:

Let I = \(\int_{0}^{\frac{\pi}{2}}\) \(\frac{\sin x}{1+\cos ^{2} x}\) dx

Put cos x = t so that – sin x dx = dt.

When x = 0, t = cos 0 = 1 and when x = \(\frac{π}{2}\), t = cos \(\frac{π}{2}\) = 0.

Question 6.

\(\int_{0}^{2}\) \(\frac{d x}{x+4-x^{2}}\)

Solution:

Question 7.

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

Solution:

Let

Question 8.

\(\int_{1}^{2}\) (\(\frac{1}{x}\) – \(\frac{1}{2 x^{2}}\) e^{2x} dx

Solution:

Choose the correct answers in questions 9 and 10:

Question 9.

The value of the integral \(\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\) dx is

(A) 6

(B) 0

(C) 3

(4) 4

Solution:

∴ Part (A) is the correct answer.

Question 10.

If f(x) = \(\int_{0}^{x}\) tsint dt, then f'(x) is

(A) cos x + x sin x

(B) x sin x

(C) x cos x

(D) sin x + x cos x

Solution:

R.H.S. = f(x) = \(\int_{0}^{x}\) tsint dt

Integrating, by parts taking t as first function, we get

∴ f'(x) = – [1.cos x – x sin x] + cos x = x sin x.

∴ Part (B) is the correct answer.