Effective ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.1 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.1

Very short answer type questions :

Evaluate the following integrals :

Question 1.
(i) ∫ x5 dx
(i) ∫ $$\frac{1}{x^3}$$ dx
(iii) ∫ $$\frac{1}{\sqrt{y}}$$ dy
Solution:
(i) ∫ x5 dx = $$\frac{x^{5+1}}{5+1}$$ + C
[∵ ∫ x5 dx = $$\frac{x^{n+1}}{n+1}$$ + C ; n ≠ 1]
= $$\frac{x^6}{6}$$ + C

(ii) ∫ $$\frac{1}{x^3}$$ dx
= ∫ x– 3 dx
= $$\frac{x^{-3+1}}{-3+1}$$ + C
= $$\frac{x^{-2}}{-2}$$ + C
= – $$\frac{1}{2 x^2}$$ + C

(iii) ∫ $$\frac{1}{\sqrt{y}}$$ dy
= ∫ y$$\frac{-1}{2}$$ dy
= $$\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$ + C
= 2√y + C.

Question 2.
(i) ∫ $$\frac{1}{\sqrt[3]{x}}$$ dx
(ii) ∫ sin2 x cosec2 x dx
(iii) ∫ sin x sec2 x dx
Solution:
(i) ∫ $$\frac{1}{\sqrt[3]{x}}$$ dx
= ∫ x– 1/3 dx
= $$\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}$$ + C
= $$\frac{3}{2}$$ x2/3 + C

(ii) ∫ sin2 x cosec2 x dx
= ∫ sin2 x × $$\frac{1}{\sin ^2 x}$$ dx
= ∫ 1 dx = x + C

(iii) ∫ sin x sec2 x dx
= ∫ $$\frac{\sin x}{\cos ^2 x}$$ dx
= ∫ tan x sec x dx
= sec x + C
[∵ ∫ tan x sec x dx = sec x + C]

Question 3.
(i) ∫ 5x dx
(ii) ∫ 7x ex dx
(iii) ∫ $$\frac{4^x}{9^x}$$ dx
Solution:
(i) ∫ 5x dx
= $$\frac{5^x}{\log _e 5}$$ + C
[∵ ∫ ax dx
= $$\frac{a^x}{\log _e a}$$ + C

(ii) ∫ 7x ex dx
= ∫ (7e)x dx
= $$\frac{(7 e)^x}{\log 7 e}$$ + C

(iii) ∫ $$\frac{4^x}{9^x}$$ dx
= ∫ $$\left(\frac{4}{9}\right)^x$$ dx
= $$\frac{\left(\frac{4}{9}\right)^x}{\log \frac{4}{9}}$$ dx + C

Question 4.
(i) ∫ $$\frac{\sin x}{1-\sin ^2 x}$$ dx
(ii) ∫ $$\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}$$ dx (NCERT)
(iii) ∫ $$\sqrt{\frac{1}{2}(1+\cos 2 x)}$$ dx
Solution:
(i) ∫ $$\frac{\sin x}{1-\sin ^2 x}$$ dx
= ∫ $$\frac{\sin x}{\cos ^2 x}$$ dx
= ∫ tan x sec x dx
= sec x + C

(ii) ∫ $$\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}$$ dx
= ∫ $$\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}$$ dx
[∵ cos 2θ = 1 – 2 sin2 θ]
= ∫ $$\frac{1}{\cos ^2 x}$$ dx
= ∫ sec2 x dx = tan x
[∵ ∫ sec2 x dx = tan x + C]

(iii) ∫ $$\sqrt{\frac{1}{2}(1+\cos 2 x)}$$ dx
= ∫ $$\sqrt{\frac{1}{2} \times 2 \cos ^2 x}$$ dx
= ∫ cos x dx
= sin x + C

Question 5.
(i) ∫ e2 log x x– 3 dx
(ii) ∫ 52 log5 x dx
(iii) ∫ 2log4 x dx
Solution:
(i) ∫ e2 log x x– 3 dx
= ∫ elog x2 x– 3 dx
= ∫ x2 . x– 3 dx
= ∫ x– 1 dx
= ∫ $$\frac{1}{x}$$ dx
= log |x| + C
[∵ ∫ $$\frac{1}{x}$$ dx = log |x| + C]

(ii) ∫ 52 log5 x dx
= ∫ 5log5 x2 dx
= ∫ x2 dx
= $$\frac{x^{3}}{3}$$ + C

(iii) ∫ 2log4 x dx
= ∫ 2$$\frac{\log x}{2 \log 2}$$ dx
= ∫ $$2^{\frac{1}{2} \log _2 x}$$ dx
= ∫ 2log2 x1/2 dx
= ∫ x$$\frac{1}{2}$$ dx
= $$\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}$$ + C
= $$\frac{2}{3}$$ x$$\frac{3}{2}$$ + C
[∵ ∫ xn dx = $$\frac{x^{n+1}}{n+1}$$ + C]