OP Malhotra Class 12 Maths Solutions Chapter 14 Indefinite Integral-2 Ex 14(a)

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S Chand Class 12 ICSE Maths Solutions Chapter 14 Indefinite Integral-2 Ex 14(a)

Question 1.
\(\int \frac{6 x-8}{3 x^2-8 x+5} d x\)
Solution:
Let I = \(\int \frac{6 x-8}{3 x^2-8 x+5} d x\)
Put 3x² – 8x + 5 = t ⇒(6x – 8) dx = dt
= \(\int \frac{d t}{t}\) = log |t| + c
= log|3x² – 8x + 5| + c

Question 2.
\(\int \frac{d x}{(3-5 x)}\)
Solution:
\(\int \frac{d x}{3-5 x}\); Put 3 – 5x = t
⇒ -5dx = dt
= \(\int \frac{d t}{-5 t}\) = \(\frac{-1}{5}\)log|t| + c
= –\(\frac{1}{5}\)log|3 – 5x| + c

Question 3.
\(\int \sqrt{1+x} d x\)
Solution:
\(\int \sqrt{1+x} d x\) = \(\int(1+x)^{\frac{1}{2}} d x\)
\(=\frac{(1+x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\)
\(=\frac{2}{3}(1+x)^{\frac{3}{2}}+c\)

Question 4.

Question 5.
\(\int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x\)
Solution:
Let I = \(\int \frac{(\cos x-\sin x) d x}{\cos x+\sin x}\)
Put cos x + sin x = t
⇒ (c – sin x + cos x) dx = dt
∴ I = \(\int \frac{d t}{t}\) = log | t | + c
= log |cos x + sin x| + c

Question 6.
\(\int \sec x \log (\sec x+\tan x) d x\)
Solution:
Let I = \(\int \sec x \log (\sec x+\tan x) d x\)
Put log (sec x + tan x) = t;
Differentate both sides w.r.t. x, we have
⇒ \(\frac{1}{\sec x+\tan x}\)(sec x tan x + sec² x)dx = dt
⇒ \(\frac{\sec x(\sec x+\tan x) d x}{\sec x+\tan x} d x\) = dt
⇒ sec x dx = dt
∴ I = \(\int t . d t\) = \(\frac{t^2}{2}\) + c
= \(\frac{1}{2}\)[log(sec x + tan x]² + c

Question 7.
\(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\)
Solution:
Put \(\sqrt{x}\) = t ⇒ \(\frac{1}{2 \sqrt{x}} d t\) = dt
∴ I = \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) = \(\int \cos t(2 d i)\)
= 2 sin t + c
= 2 sin \(\sqrt{x}\) + c

Question 8.
\(\int x e^{x^2} d x\)
Solution:
Let I = \(\int x e^{x^2} d x\);
Put x² = t ⇒ 2xdx = dt
= \(\int e^t \frac{d t}{2}=\frac{1}{2} e^t+c\)
= \(\frac{1}{2} e^{x^2}+c\)

Question 9.
\(\int \frac{e^{m \tan ^{-1} x}}{1+x^2} d x\)
Solution:
Let I = \(\int \frac{e^{m \tan ^{-1} x}}{1+x^2} d x\)
Put tan^-1 x = t ⇒ \(\frac{1}{1+x^2} d x\) = dt
∴ I = \(\int e^{m t} d t\) = \(\frac{e^{m t}}{m}+c\)
= \(\frac{e^{m \tan ^{-1}}}{m}+c\)

Question 10.
\(\int \frac{\sec ^2(\log x)}{x} d x\)
Solution:
Let I = \(\int \frac{\sec ^2(\log x)}{x} d x\)
Put log x = t ⇒ \(\frac { 1 }{ x }\)dx = dt
∴ I = \(\int \sec ^2 t d t\) = tan t + c
= tan (log x) + c

Question 11.
\(\int \sin ^2\)x cos x dx
Solution:
Let I = \(\int \sin ^2\)x cos x dx
Put sin x = t ⇒ cos x dx = dt
= \(\int t^2 d t\) = \(\frac{t^3}{3}\) + c
= \(\frac{\sin ^3 x}{3}+c\)

Question 12.
\(\int \frac{\left(\sin ^{-1} x\right)^2}{\sqrt{\left(1-x^2\right)}} d x\)
Solution:
Let I = \(\int \frac{\left(\sin ^{-1} x\right)^3}{\sqrt{1-x^2}} d x\)
\(=\int\left(\sin ^{-1} x\right)^3 \cdot \frac{1}{\sqrt{1-x^2}} d x\)
\(=\frac{\left(\sin ^{-1} x\right)^4}{4}+c\)

Question 13.
\(\int \sec ^4\)x tan x dx
Solution:
Let I = \(\int \sec ^4\)x tan x dx
= \(\int \sec ^3\)x (sec x tan x dx)
Put sec x = t ⇒ sec x tan x dx = dt
= \(\int t^3 d t\) = \(\frac{t^4}{4}+c\)
= \(\frac{\sec ^4 x}{4}+c\)

Question 14.
\(\int \frac{x^3}{\left(x^2+1\right)^3} d x\)
Solution:

Question 15.
\(\int \frac{1+\tan x}{x+\log \sec x} d x\)
Solution:
Let I = \(\int \frac{1+\tan x}{x+\log \sec x} d x\)
Put x + log sec x = t; on differentiating
\(=\int t^3 d t=\frac{t^4}{4}+c\)
\(=\frac{\sec ^4 x}{4}+c\)

Question 16.
\(\int \sec ^4\)x dx
Solution:
\(\int \sec ^4\)x dx = \(\int \sec ^2\)(1 + tan² x) dx
= \(\int\left(1+t^2\right) d t\) = \(t+\frac{t^3}{3}+c\)
= tan x + \(\frac{\tan ^3 x}{3}+c\)

Question 17.
\(\int \frac{\sin (\log x)}{x} d x\)
Solution:
Let I = \(\int \frac{\sin (\log x)}{x}\)dx;
Solution:
Put log x = t ⇒ \(\frac { 1 }{ x }\)dx = dt
= \(\int(\sin t d t)\) = -cos t + c
= – cos(log x) + c

Question 18.
\(\int \frac{\cos x-\sin x}{1+\sin 2 x} d x\)
Solution:

Question 19.
\(\int\left(\frac{1+\sin x}{1+\cos x}\right) d x\)
Solution:

Question 20.
\(\int \frac{(\sin \theta+\cos \theta)}{\sqrt{\sin 2 \theta}} d \theta\)
Solution:
Let I = \(\int \frac{(\sin \theta+\cos \theta) d \theta}{\sqrt{\sin 2 \theta}}\)
Put sin θ – cos θ = t …(1)
⇒ (c0s θ + sin θ) dθ = dt
On squaring eqn (1) both sides; we have (sin θ – cos θ)² = t²
⇒ sin²θ + cos 2θ – 2 sin θ cos θ = t²
⇒ 1 – sin 2θ = t²
⇒ sin 2θ = 1 – t²

Question 21.

Question 22.
\(\int \frac{\sin x \cos x d x}{\sin ^4 x+\cos ^4 x}\)
Solution:

Question 23.
\(\int \frac{\tan x}{\sec x+\tan x} d x\)
Solution:
Let I = \(\int \frac{\tan x d x}{\sec x+\tan x}\)
= \(\int \frac{\tan x(\sec x-\tan x) d x}{\sec ^2 x-\tan ^2 x}\)
= \(\int \tan x\) sec x dx – \(\int \tan ^2\)x dx
\(=\int \tan x \sec x d x-\int\left(\sec ^2 x-1\right) d x\)
= sex x – tan x + x + c

Question 24.
\(\int \tan ^4 x d x\)
Solution:

Question 25.
\(\int \frac{d x}{\sqrt{1+\sqrt{x}}} d x\)
Solution:

Question 26.
\(\int \frac{x d x}{\sqrt{x+5}}\)
Solution:

Question 27.
\(\int \frac{(x+1)(x+\log x)^2}{x} d x\)
Solution:

Question 28.
\(\int x^2 e^{x^3} \cos \left(e^{x^3}\right) d x\)
Solution:
Let I = \(\int x^2 e^{x^3} \cos \left(e^{x^3}\right) d x\)
Put \(e^{x^3}=t \Rightarrow e^{x^3} \times 3 x^2 d x=d t\)
= \(\int \cos t \frac{d t}{3}=\frac{1}{3} \sin t+c\)
= \(\frac{1}{3} \sin \left(e^{x^3}\right)+c\)

Question 29.
\(\int \frac{2 \sin \theta \cos \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta\)
Solution:
Let I = \(\int \frac{2 \sin \theta \cos \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta\)
Divide numerator and denominator by cos⁴ θ; we have

Question 30.
\(\int \frac{1}{9+16 \cos ^2 x} d x\)
Solution:
Let I = \(\int \frac{1}{9+16 \cos ^2 x} d x\);
Divide numerator and denominator by cos² x; we have

Question 31.
\(\int \frac{\sin x}{a+b \cos x} d x\)
Solution:

Question 32.
\(\int \frac{e^x-1}{e^x+1} d x\)
Solution:

Question 33.
\(\int \frac{e^{2 x}}{e^{2 x}-2} d x\)
Solution:

Question 34.
\(\int \frac{e^{2 x}}{e^x-1} d x\)
Solution:
Let I = \(\int \frac{e^{2 x} d x}{e^x-1}\)
Put e^x – 1 = t ⇒ e^x = t + 1
⇒ e^x dx = dt
= \(\int \frac{(t+1) d t}{t}\) = \(\int\left(1+\frac{1}{t}\right) d t\)
= t + log | t | + e
= e^x + log | e^x-1 | + c’ [where c’ = c – 1]

Question 35.
\(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)
Solution:
Let I = \(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)
Put e^2x + e^-2x = t ⇒ (2e^2x – 2e^-2x) dx = dt
⇒ (e^2x – e^-2x) dx = \(\frac{d t}{2}\)
= \(\int \frac{d t}{2 t}\) = \(\frac { 1 }{ 2 }\)log |t| + c
= \(\frac { 1 }{ 2 }\)log|e^2x + e^-2x | + c

Question 36.

Put cot x = t ⇒ -cosec²x dx = dt
= \(-\int t^3\left(1+t^2\right) d t\)
= –\(\frac{t^4}{4}\) –\(\frac{t^4}{4}\) + c
= – \(\frac { 1 }{ 4 }\)cot⁴x – \(\frac { 1 }{ 6 }\)cot⁶x + c

Question 37.
\(\int \sqrt{e^x-4} d x\)
Solution:
Let I = \(\int \sqrt{e^x-4} d x\)
Put \(\sqrt{e^x-4}=t\); on squaring both sides
e^x – 4 = t² ⇒ e^x = t² + 4 ⇒ e^x dx = 2t dt

Question 38.
\(\int \frac{x d x}{e^{x^2}}\)
Solution:

Question 39.
Evaluate:

Question 40.
\(\int\left(x^3-1\right)^{1 / 3} x^5 d x\)
Solution:

Question 41.
\(\int \sin ^3 x \cos ^4 x d x\)
Solution:
Let I = \(\int \sin ^3 x \cos ^4 x d x=\int \sin x\left(1-\cos ^2 x\right) \cos ^4 x d x\)
Put cos x = t ⇒ – sin x dx = dt = – \(-\int\left(1-t^2\right) t^4 d t\)
= – \(\frac{t^5}{5}\) + \(\frac{t^7}{7}\) + C = – \(\frac{\cos ^5 x}{5}\) + \(\frac{\cos ^7 x}{7}\) + C