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

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 e^x = t
e^x 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