Well-structured Understanding ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.8 facilitate a deeper understanding of mathematical principles.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.8

Evaluate the following (1 to 12) integrals:

Question 1.
(i) ∫ $$\frac{d x}{x^2-4 x+8}$$
(ii) ∫ $$\frac{d x}{x^2+2 x+2}$$
Solution:
(i) ∫ $$\frac{d x}{x^2-4 x+8}$$
= ∫ $$\frac{d x}{x^2-4 x+4+4}$$
= ∫ $$\frac{d x}{(x-2)^2+2^2}$$
[put x – 2 = t
⇒ dx = dt]
= ∫ $$\frac{d t}{t^2+a^2}$$
= $$\frac{1}{2} \tan ^{-1} \frac{t}{2}$$ + C
[∵ ∫ $$\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$]
= $$\frac{1}{2} \tan ^{-1}\left(\frac{x-2}{2}\right)$$ + C

(ii) ∫ $$\frac{d x}{x^2+2 x+2}$$
Solution:
Let I = ∫ $$\frac{d x}{x^2+2 x+2}$$
= ∫ $$\frac{d x}{x^2+2 x+1+1}$$
= ∫ $$\frac{d x}{(x+1)^2+1^2}$$
= ∫ $$\frac{d t}{t^2+1^2}$$ ;
when x + 1 = t
⇒ dx = dt
= $$\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)$$ + C
= tan-1 (x + 1) + C

Question 1 (old).
(ii) ∫ $$\frac{d x}{5-8 x-x^2}$$ (NCERT)
Solution:
Let I = ∫ $$\frac{d x}{5-8 x-x^2}$$

Question 2.
(i) ∫ $$\frac{d x}{x^2-6 x+13}$$ (NCERT)
(ii) ∫ $$\frac{x+3}{x^2-2 x-5}$$ dx (NCERT)
Solution:
(i) ∫ $$\frac{d x}{x^2-6 x+13}$$
= ∫ $$\frac{d x}{x^2-6 x+9+4}$$
= ∫ $$\frac{d x}{(x-3)^2+2^2}$$
put x – 3 = t
dx = dt
= ∫ $$\frac{d t}{t^2+2^2}$$
= $$\frac{1}{2} \tan ^{-1} \frac{t}{2}$$ + C
= $$\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)$$ + C

(ii) Let I = ∫ $$\frac{x+3}{x^2-2 x-5}$$ dx

Question 3.
(i) ∫ $$\frac{x+2}{2 x^2+6 x+5}$$ dx
(ii) ∫ $$\frac{2 x+1}{18-4 x-x^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x+2}{2 x^2+6 x+5}$$ dx

(ii) Let I = ∫ $$\frac{2 x+1}{18-4 x-x^2}$$ dx
Let 2x + 1 = A $$\frac{d}{d x}$$ (18 – 4x – x2) + B
= A (- 4 – 2x) + B
On comparing the coefficient of x and constant terms on both sides ; we get
2 = – 2 A
⇒ A = – 1 ;
1 = – 4A + B
B = – 3

Question 4.
(i) ∫ $$\frac{e^x}{2 e^{2 x}+3 e^x+5}$$ dx
(ii) ∫ $$\frac{\cos x}{6+4 \sin x-\cos ^2 x}$$ dx
Solution:
(i) ∫ $$\frac{e^x}{2 e^{2 x}+3 e^x+5}$$ dx ;
put ex = t
ex dx = dt
∴ I = ∫ $$\frac{d t}{2 t^2+3 t+5}$$
= $$\frac{1}{2}$$ ∫ $$\frac{d t}{t^2+\frac{3 t}{2}+\frac{5}{2}}$$
= $$\frac{1}{2}$$ ∫ $$\frac{d t}{t^2+\frac{3 t}{2}+\frac{9}{16}-\frac{9}{16}+\frac{5}{2}}$$

(ii) Let I = ∫ $$\frac{\cos x}{6+4 \sin x-\cos ^2 x}$$ dx

= tan-1 (t + 2) + C
= tan-1 (2 + sin x) + C

Question 5.
(i) ∫ $$\frac{d x}{\sqrt{2 x-x^2}}$$ (NCERT)
(ii) ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$
Solution:
(i) ∫ $$\frac{d x}{\sqrt{2 x-x^2}}$$
= ∫ $$\frac{d x}{\sqrt{-\left(x^2-2 x+1-1\right)}}$$
= ∫ $$\frac{d x}{\sqrt{1-(x-1)^2}}$$
put x – 1 = t
⇒ dx = dt
= ∫ $$\frac{d t}{\sqrt{1-t^2}}$$
= sin-1 t + C
= sin-1 (x – 1) + C
[∵ ∫ $$\frac{d x}{\sqrt{a^2-x^2}}$$ = sin-1 $$\frac{x}{a}$$ + C]

(ii) Let I = ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$

Question 5 (old).
(ii) ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$
Solution:
Let I = ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$
= ∫ $$\frac{d x}{\sqrt{x^2-3 x+\frac{9}{4}-\frac{9}{4}+2}}$$
= ∫ $$\frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}$$
put x – $$\frac{3}{2}$$ = t
⇒ dx = dt
= ∫ $$\frac{d t}{\sqrt{t^2-\left(\frac{1}{2}\right)^2}}$$
= log |t + $$\sqrt{t^2-\frac{1}{4}}$$| + C
= log |x – $$\frac{3}{2}+\sqrt{x^2-3 x+2}$$| + C
[∵∫ $$\frac{d x}{\sqrt{x^2-a^2}}$$ = log |x + $$\sqrt{x^2-a^2}$$| + C|]

Question 6.
(i) ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$ (NCERT)
(ii) ∫ $$\frac{d x}{\sqrt{(x-1)(x-2)}}$$ (NCERT)
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{7-6 x-x^2}}$$

(ii) Let I = ∫ $$\frac{d x}{\sqrt{(x-1)(x-2)}}$$

Question 6 (old).
(ii) ∫ $$\frac{d x}{\sqrt{x^2+2 x+2}}$$ (NCERT)
Solution:
Let I = ∫ $$\frac{d x}{\sqrt{x^2+2 x+2}}$$
= ∫ $$\frac{d x}{\sqrt{x^2+2 x+1+1}}$$
= ∫ $$\frac{d x}{\sqrt{(x+1)^2+1^2}}$$
put x + 1 = t
⇒ dx = dt
= ∫ $$\frac{d t}{\sqrt{t^2+1^2}}$$
= log |t + $$\sqrt{t^2+1}$$| + C
[∵∫ $$\frac{d x}{\sqrt{x^2-a^2}}$$ = log |x + $$\sqrt{x^2-a^2}$$| + C|]
= log |x + 1 + $$\sqrt{x^2+2 x+2}$$| + C

Question 7.
(i) ∫ $$\frac{d x}{\sqrt{5-4 x-2 x^2}}$$
(ii) ∫ $$\frac{d x}{\sqrt{5 x-4 x^2}}$$ (ISC 2020)
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{5-4 x-2 x^2}}$$
= ∫ $$\frac{d x}{\sqrt{-2\left(x^2+2 x-\frac{5}{2}\right)}}$$
= ∫ $$\frac{d x}{\sqrt{2} \sqrt{-\left(x^2+2 x+1-1-\frac{5}{2}\right)}}$$
= $$\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{-(x+1)^2+\frac{7}{2}}}$$
= $$\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2-(x+1)^2}}$$
= $$\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+1}{\sqrt{\frac{7}{2}}}\right)$$ + c
= $$\frac{1}{\sqrt{2}} \sin ^{-1}\left[\sqrt{\frac{2}{7}}(x+1)\right]$$ + c

(ii) Let I = ∫ $$\frac{d x}{\sqrt{5 x-4 x^2}}$$

Question 7 (old).
(ii) ∫ $$\frac{d x}{\sqrt{8+3 x-x^2}}$$
Solution:
Let I = ∫ $$\frac{d x}{\sqrt{8+3 x-x^2}}$$
= ∫ $$\frac{d x}{\sqrt{-\left(x^2-3 x-8\right)}}$$
= ∫ $$\frac{d x}{\sqrt{-\left(x^2-3 x+\frac{9}{4}-\frac{9}{4}-8\right)}}$$
= ∫ $$\frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^2}}$$
= ∫ $$\frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}$$
= sin-1 $$\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)$$ + c
= sin-1 $$\left(\frac{2 x-3}{\sqrt{41}}\right)$$ + c
[∵∫ $$\frac{d x}{\sqrt{a^2-x^2}}$$ = sin-1 $$\frac{x}{a}$$ + c]

Question 8.
(i) ∫ $$\frac{x-1}{\sqrt{x^2-x}}$$ dx (NCERT)
(ii) ∫ $$\frac{5 x+3}{\sqrt{x^2+4 x+10}}$$ dx
Solution:
(i) Let x – 1 = A $$\frac{d}{d x}$$ (x2 – x) + B
= A (2x – 1) + B
⇒ x – 1 = 2Ax – A + B
∴ 2A = 1
A = $$\frac{1}{2}$$
and – A + B = – 1
⇒ B = – 1 + $$\frac{1}{2}$$ = – $$\frac{1}{2}$$
∴ I = ∫ $$\frac{x-1}{\sqrt{x^2-x}}$$ dx
= ∫ $$\frac{\frac{1}{2}(2 x-1)-\frac{1}{2}}{\sqrt{x^2-x}}$$ dx
= $$\frac{1}{2} \int \frac{(2 x-1) d x}{\sqrt{x^2-x}}-\frac{1}{2} \int \frac{d x}{\sqrt{x^2-x}}$$
= $$\frac{1}{2}$$ I1 – $$\frac{1}{2}$$ I2
where I1 = $$\frac{(2 x-1) d x}{\sqrt{x^2-x}}$$
put x2 – x = t
⇒ (2x – 1) dx = dt

(ii) Let I = ∫ $$\frac{5 x+3}{\sqrt{x^2+4 x+10}}$$ dx

Question 8 (old).
(i) ∫ $$\frac{d x}{\sqrt{5 x^2-2 x}}$$ (NCERT)
(ii) ∫ $$\frac{d x}{\sqrt{9 x-4 x^2}}$$ (NCERT)
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{5 x^2-2 x}}$$
= ∫ $$\frac{d x}{\sqrt{5\left(x^2-\frac{2}{5} x+\frac{1}{25}-\frac{1}{25}\right)}}$$
= $$\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}$$
= $$\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}\right|$$ + c
= $$\frac{1}{\sqrt{5}} \log \left|\frac{5 x-1}{5}+\frac{\sqrt{5 x^2-2 x}}{\sqrt{5}}\right|$$ + c
[∵∫ $$\frac{d x}{\sqrt{a^2-x^2}}$$ = log |x + $$\sqrt{x^2-a^2}$$| + c]

(ii) Let I = ∫ $$\frac{d x}{\sqrt{9 x-4 x^2}}$$

Question 9.
(i) ∫ $$\frac{3 x+5}{\sqrt{x^2-8 x+7}}$$ dx
(ii) ∫ $$\frac{x^2}{\sqrt{x^6-2 x^2-3}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{3 x+5}{\sqrt{x^2-8 x+7}}$$ dx

∴ from (1) ; we have
I = 3 $$\sqrt{x^2-8 x+7}$$ + 17 log |x – 4 + $$\sqrt{x^2-8 x+7}$$| + C

(ii) Let I = ∫ $$\frac{x^2}{\sqrt{x^6-2 x^2-3}}$$ dx ;
put x3 = t
⇒ 3x2 dx = dt
∴ I = ∫ $$\frac{d t}{3 \sqrt{t^2-2 t-3}}$$
= $$\frac{1}{3} \int \frac{d t}{\sqrt{t^2-2 t+1-4}}$$
= $$\frac{1}{3} \int \frac{d t}{\sqrt{(t-1)^2-2^2}}$$
put t – 1 = u
⇒ dt = du
= $$\frac{1}{3} \int \frac{d u}{\sqrt{u^2-2^2}}$$
= $$\frac{1}{3}$$ log |u + $$\sqrt{u^2-4}$$| + C
[∵∫ $$\frac{d x}{\sqrt{x^2+^2}}$$ = log |x + $$\sqrt{x^2-a^2}$$| + c]
= $$\frac{1}{3}$$ log |t – 1 + $$\sqrt{t^2-2 t-3}$$|
= $$\frac{1}{3}$$ log |x – 1 + $$\sqrt{x^6-2 x^3-3}$$| + C

Question 10 (old).
(i) ∫ $$\frac{x+3}{\sqrt{5-4 x-x^2}}$$ dx
Solution:
Let I = $$\frac{(x+3) d x}{\sqrt{5-4 x-x^2}}$$

Question 11 (old).
(ii) ∫ $$\frac{x+2}{\sqrt{(x-2)(x-3)}}$$ dx
Solution:
Let I = ∫ $$\frac{x+2}{\sqrt{(x-2)(x-3)}}$$ dx

Question 12 (old).
(i) ∫ $$\frac{x+2}{\sqrt{x^2+2 x+3}}$$ dx
Solution:
Let I = ∫ $$\frac{x+2}{\sqrt{x^2+2 x+3}}$$ dx