ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.17

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.17

Evaluate the following (1 to 21) definite integrals :

Question 1.
(i) \(\int_{-1}^1\) x² (x³ + 1)³ dx
(ii) \(\int_0^1\) x e^x2 dx (NCERT)
Solution:
(i) Let I = \(\int_{-1}^1\) x² (x³ + 1)³ dx
put x³ = t
⇒ 3x² dx = dt
When x = – 1 ⇒ t = – 1
When x = 1 ⇒ t = 1
∴ I = \(\int_{-1}^1\) (t + 1)³ \(\frac{d t}{3}\)
= \(\left.\frac{1}{3} \frac{(t+1)^4}{4}\right]_{-1}^1\)
= \(\frac{1}{12}\) [2⁴ – 0⁴]
= \(\frac{16}{12}\)
= \(\frac{4}{3}\)

(ii) Let I = \(\int_0^1\) x e^x2 dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1² = 1
∴ I = \(\int_0^1 e^t \frac{d t}{2}\)
= \(\frac{1}{2}\left[e^t\right]_0^1\)
= \(\frac{1}{2}\) (e – 1).

Question 2.
(i) \(\int_0^1 \frac{x}{x^2+1}\) dx
(ii) \(\int_0^{\pi / 3} \frac{\cos x}{3+4 \sin x}\) dx
Solution:
(i) Let I = \(\int_0^1 \frac{x}{x^2+1}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0
and When x – 1 ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2(t+1)}\)
= \(\left.\frac{1}{2} \log |t+1|\right]_0^1\)
= \(\frac{1}{2}\) log 2

(ii) Let I = \(\int_0^{\pi / 3} \frac{\cos x}{3+4 \sin x}\) dx
put sin x = t
⇒ cos x dx =dt
When x = 0 ⇒ t = 0
When x = \(\frac{\pi}{3}\)
⇒ t = \(\frac{\sqrt{3}}{2}\)

Question 3.
(i) \(\int_1^2 \frac{3 x}{9 x^2-1}\) dx
(ii) \(\int_a^b \frac{\log x}{x}\) dx
(iii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx
Solution:
(i) Let I = \(\int_1^2 \frac{3 x}{9 x^2-1}\) dx

(ii) Let I = \(\int_a^b \frac{\log x}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = a
⇒ t = log a
When x = b
⇒ t = log b

(iii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx

Question 4.
(i) \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx (NCERT)
(ii) \(\int_0^{\pi / 2} \frac{\sin x}{\sqrt{1+\cos x}}\) dx
Solution:
(i) Let I = \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx
= \(\left.\int_0^2 \frac{6 x d x}{x^2+4}+\frac{3}{2} \tan ^{-1} \frac{x}{2}\right]_0^2\)
put x² = t in Ist integral
∴ 2x dx = dt
When x = 0 ⇒ t = 0
and When x = 2 ⇒ t = 4

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin x}{\sqrt{1+\cos x}}\) dx

Question 5.
(i) \(\int_0^1 \frac{x^5}{1+x^6}\) dx
(ii) \(\int_0^1 \frac{x}{\sqrt{1+x^2}}\) (NCERT Exemplar)
Solution:
(i) Let I = \(\int_0^1 \frac{x^5}{1+x^6}\) dx
put x⁶ = t
⇒ 6x⁵ dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1
∴ I = \(\left.\int_0^1 \frac{d t}{6(1+t)}=\frac{1}{6} \log (1+t)\right]_0^1\)
= \(\frac{1}{6}\) [log 2 – log 1]
= \(\frac{1}{6}\) log 2

(ii) Let I = \(\int_0^1 \frac{x dx}{\sqrt{1+x^2}}\)

Question 6.
(i) \(\int_1^e \frac{1+\log x}{2 x}\) dx
(ii) \(\int_0^1 \frac{x}{1+x^4}\) dx
Solution:
(i) Let I = \(\int_1^e \frac{1+\log x}{2 x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = log 1 = 0
When x = e
⇒ t = log e = 1
= \(\int_0^1 \frac{(1+t)}{2} d t\)
= \(\left.\frac{1}{2} \frac{(1+t)^2}{2}\right]_0^1\)
= \(\frac{1}{4}\) [2² – 1]
= \(\frac{3}{4}\)

(ii) Let I = \(\int_0^1 \frac{x}{1+x^4}\) dx
put x² dx = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2\left(1+t^2\right)}\)
= \(\left.\frac{1}{2} \tan ^{-1} t\right]_0^1\)
= \(\frac{1}{2}\) [tan^-1 1 – tan^-1 0]
= \(\frac{1}{2} \times \frac{\pi}{4}=\frac{\pi}{8}\)

Question 6 (old).
(i) \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
(ii) \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)
Solution:
(i) Let I = \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)

(ii) Let I = \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)

Question 7.
(i) \(\int_0^1 \frac{e^{2 x}}{1+e^{4 x}}\) dx
(ii) \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin³ x dx
Solution:
(i) Let I = \(\int_0^1 \frac{e^{2 x}}{1+e^{4 x}}\) dx
put e^2x = t
⇒ 2 e^2x dx = dt
When x = 0
⇒ t = 1 ;
When x = 1
⇒ t = e²

(ii) Let I = \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin³ x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) (1 – cos² x) sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 0

i

Question 7 (old).
(i) \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\) (NCERT)
Solution:
Let I = \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)

Question 8.
(i) \(\int_0^{\pi / 4}\) sin³ 2x cos 2x dx
(ii) \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁵ x dx
Solution:
(i) Let I = \(\int_0^{\pi / 4}\) sin³ 2x cos 2x dx
put sin 2x = y
⇒ + 2 cos 2x dx = dy
When x = 0 ⇒ y = 0
and when x = \(\frac{\pi}{4}\)
⇒ y = sin \(\frac{\pi}{2}\) = 1
= \(\int_0^1 y^3\left(+\frac{d y}{2}\right)\)
= \(+\frac{1}{2}\left[\frac{y^4}{4}\right]_0^1\)
= + \(\frac{1}{2}\left(\frac{1}{4}-0\right)\)
= + \(\frac{1}{8}\)

(ii) Let I = \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁵ x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁴ x . sin x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) (1 – cos² x)² . sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0
⇒ t = 1
When x = \(\frac{\pi}{2}\)
⇒ t = 0

Question 9.
(i) \(\int_1^3 \frac{\cos (\log x)}{x}\) dx
(ii) \(\int_1^2 \frac{d x}{x(1+\log x)^2}\)
Solution:
(i) Let I = \(\int_1^3 \frac{\cos (\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = log 1 = 0
and when x = 3
⇒ t = log 3
∴ I = \(\int_0^{\log 3}\) cos t dt
= sin t \(]_0^{\log 3}\)
= sin (log 3) – sin 0
= sin (log 3) – 0
= sin (log 3)

(ii) Let I = \(\int_1^2 \frac{d x}{x(1+\log x)^2}\)
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1 ⇒ t = 0
and When x = 2 ⇒ t = log 2
= \(\left.\int_0^{\log 2} \frac{d t}{(1+t)^2}=-\frac{1}{1+t}\right]_0^{\log 2}\)
= – \(\frac{1}{1+\log 2}\) + 1
= \(\frac{-1+1+\log 2}{1+\log 2}\)
= \(\frac{\log 2}{1+\log 2}\)

Question 10.
(i) \(\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x}\) dx
(ii) \(\int_0^1\) tan^-1 x dx
Solution:
(i) Let I = \(\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x}\) dx ;
put sec x = t
⇒ sec x tan x dx = dt
When x = 0 ⇒ t = 1
and When x = \(\frac{\pi}{3}\) ⇒ t = 2
∴ I = \(\int_1^2 \frac{d t}{\left(1+t^2\right)}\)
= tan^-1 t\(]_1^2\)
= tan^-1 2 – tan^-1 1
Thus, I = tan^-1 2 – \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^1\) tan^-1 x . 1 dx

Question 10 (old).
(ii) \(\int_4^8 x \sqrt[3]{x-4}\) dx
Solution:
Let I = \(\int_4^8 x \sqrt[3]{x-4}\) dx
put x – 4 = t
⇒ dx = dt
When x = 4 ⇒ t = 0 ;
When x = 8 ⇒ t = 4

Question 11.
(i) \(\int_0^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
(ii) \(\int_0^1\) sin^-1 x dx (NCERT)
Solution:
(i) Let I = \(\int_0^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
put sin^-1 x = t
⇒ x = sin t
⇒ dx = cos t dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{1}{2}\) ⇒ t = \(\frac{\pi}{6}\)

(ii) Let I = \(\int_0^1\) sin^-1 x dx
= \(\int_0^1\) sin^-1 x . 1 dx

Question 12.
(i) \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
(ii) \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)
Solution:
(i) Let I = \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
= \(\int_0^4 \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}\)
= log |x + 1 + \(\sqrt{x^2+2 x+3}\)|\(]_0^4\)
= log |5 + \(\sqrt{16+8+3}\)| – log |1 + √3|
= log |5 + 3√3| – log |1 + √3|
= log \(\left|\frac{5+3 \sqrt{3}}{1+\sqrt{3}}\right|\)

(ii) Let I = \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)

Question 13.
(i) \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)}\) dx
(ii) \(\int_0^1 \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)}\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)}\) dx
put sin x = t
⇒ cos x dx = dt
When x = 0 ⇒ t = 0 and
When x = \(\frac{\pi}{2}\) ⇒ t = 1

(ii) Let I = \(\int_0^1 \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0
When x = 1 ⇒ t = 1

Question 14.
(i) \(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x}\) dx
(ii) \(\int_0^1 \frac{d x}{2 e^x-1}\)
(iii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx
(iv) \(\int_0^a \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x}\) dx
put sin² x = t
⇒ 2 sin x cos x dx = dt
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\)
⇒ t = sin² \(\frac{\pi}{2}\) = 1

(ii) Let I = \(\int_0^1 \frac{d x}{2 e^x-1}\)
put 2e^x – 1 = t
⇒ e^x = \(\frac{t+1}{2}\)
⇒ e^x dx = \(\frac{1}{2}\) dt
When x = 0 ⇒ t = 2 – 1 = 1
When x = 1 ⇒ t = 2e – 1

(iii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx

(iv) Let I = \(\int_0^a \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\) dx
put x = a tan² θ
⇒ dx = 2a tan θ sec² θ
When x = 0 ⇒ θ = 0 ;
When x = a ⇒ θ = \(\frac{\pi}{4}\)

Question 15.
(i) \(\int_0^1 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) dx
(ii) \(\int_0^1 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) dx
Solution:
(i) put x = tan θ
⇒ dx = sec² θ dθ
When x = 0 ⇒ θ = 0
and When x = 1 ⇒ θ = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^1 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) dx
put x = tan θ
⇒ dx = sec² θ dθ
When x = 0 ⇒ tan θ = 0 ⇒ θ = 0
When x = 1 ⇒ tan θ = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 15 (old).
(i) \(\int_0^1\) tan^-1 x dx
(ii) \(\int_0^1\) x (tan^-1 x)² dx
Solution:
(i) Let I = \(\int_0^1\) tan^-1 x dx
= \(\int_0^1\) tan^-1 x . 1 dx

(ii) L.H.S. = \(\int_0^1\) x (tan^-1 x)² dx

Question 16.
(i) \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^2}\) dx
(ii) \(\int_0^{\pi / 2} \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)
Solution:
(i) Let I = \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^2}\) dx
put tan^-1 x = t
⇒ x = tan t
⇒ dx = sec² t dt
When x = 0 ⇒ t = 0 and
When x = 1 ⇒ t = tan^-1 1 = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^{\pi / 2} \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)
Divide Numerator and denominator by cos² x ; we have
I = \(\int_0^{\pi / 2} \frac{\sec ^2 x d x}{a^2 \tan ^2 x+b^2}\)
put tan x = t
⇒ sec² x dx = dt
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\)
⇒ t = tan \(\frac{\pi}{2}\) → ∞

Question 17.
(i) \(\int_0^1 \frac{5 x}{\left(4+x^2\right)^2}\) dx
(ii) \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x}\) dx (NCERT)
Solution:
(i) Let I = \(\int_0^1 \frac{5 x}{\left(4+x^2\right)^2}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1

(ii) Let I = \(\int_0^{\pi / 4} \frac{\sin x \cos x d x}{\sin ^4 x+\cos ^4 x}\)
Divide Numerator and denominator by cos⁴ x ; we get
I = \(\int_0^{\pi / 4} \frac{\tan x \sec ^2 x d x}{\tan ^4 x+1}\)
put tan² x = t
⇒ 2 tan x sec² x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{4}\) ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2\left(t^2+1\right)}\)
= \(\left.\frac{1}{2} \tan ^{-1} t\right]_0^1\)
= \(\frac{1}{2} \times \frac{\pi}{4}=\frac{\pi}{8}\)

Question 18.
(i) \(\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}\)
(ii) \(\int_0^{\pi / 2} \frac{d x}{4 \cos x+2 \sin x}\)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d x}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{2 d x}{1+t^2}\)
and sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{2 t}{1+t^2}\)
∴ When x = 0 ⇒ t = 0
and when x = \(\frac{\pi}{2}\) ⇒ t = 1

(ii) \(\int_0^{\pi / 2} \frac{d x}{4 \cos x+2 \sin x}\)

Question 19.
(i) \(\int_0^\pi \frac{d x}{5+3 \cos x}\)
(ii) \(\int_0^\pi \frac{d x}{6-\cos x}\)
Solution:
(i) Let I = \(\int_0^\pi \frac{d x}{5+3 \cos x}\)

(ii) put tan \(\frac{x}{2}\) = t
⇒ sec²\(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-t^2}{1+t^2}\)
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\) ⇒ t = 1

Question 20.
(i) \(\int_0^1 \frac{1-x^2}{x^4+x^2+1}\) dx
(ii) \(\int_0^1 \frac{x^3}{\left(1+x^2\right)^4}\) dx
Solution:
(i) Let I = \(\int_0^1 \frac{1-x^2}{x^4+x^2+1}\) dx
Divide Numerator and denominator by x² ; we have
= \(\int_0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{x^2+1+\frac{1}{x^2}}\)

(ii) Let I = \(\int_0^1 \frac{x^3}{\left(1+x^2\right)^4}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1