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 x² + 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})cos⁵ϕdϕ
Solution:
Let I = (int_{0}^{frac{pi}{2}}) (sqrt{sinϕ})cos⁵ϕdϕ
= (int_{0}^{frac{pi}{2}}) (sqrt{sinϕ})cos⁴ϕcosϕdϕ
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 = t² so that dx = 2t dt.
When x = 0, t = (sqrt{2}) and when x = 2, t² = 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.