CBSE Class 12 Mathematics Integrals Worksheet Set 02

Question. Write the antiderivative of \( \left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) \). (Delhi 2014)
Answer: The antiderivative of \( 3\sqrt{x} + \frac{1}{\sqrt{x}} \)
\( = \int \left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) dx = 3\int x^{1/2}dx + \int x^{-1/2}dx \)
\( = 3 \cdot \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C = 2x\sqrt{x} + 2\sqrt{x} + C \)
\( = 2\sqrt{x}(x + 1) + C \)

Question. Evaluate : \( \int \cos^{-1}(\sin x) dx \) (Delhi 2014)
Answer: \( \int \cos^{-1}(\sin x) dx = \int \cos^{-1}\left[ \cos\left( \frac{\pi}{2} - x \right) \right] dx \)
\( = \int \left( \frac{\pi}{2} - x \right) dx = \frac{\pi}{2}x - \frac{x^2}{2} + C \)

Question. Evaluate : \( \int \frac{dx}{\sin^2 x \cos^2 x} \) (Foreign 2014, Delhi 2014C)
Answer: We have \( \int \frac{dx}{\sin^2 x \cos^2 x} = \int \frac{(\sin^2 x + \cos^2 x)}{\sin^2 x \cos^2 x} dx \)
\( = \int (\sec^2 x + \operatorname{cosec}^2 x) dx = \tan x - \cot x + C \)

Question. Find : \( \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx \) (Delhi 2014C)
Answer: Refer to answer 3.

Question. Evaluate : \( \int (1 - x) \sqrt{x} \, dx \) (Delhi 2012)
Answer: \( \int (1 - x) \sqrt{x} \, dx = \int (x^{1/2} - x^{3/2}) dx \)
\( = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C \)

Question. Write the value of \( \int \frac{\sec^2 x}{\operatorname{cosec}^2 x} dx \) (Delhi 2012C, 2011)
Answer: \( \int \frac{\sec^2 x}{\operatorname{cosec}^2 x} dx = \int \frac{\sin^2 x}{\cos^2 x} dx = \int \tan^2 x dx \)
\( = \int (\sec^2 x - 1) dx = \tan x - x + C \)

Question. Write the value of \( \int \frac{2 - 3\sin x}{\cos^2 x} dx \) (Delhi 2011)
Answer: \( \int \frac{2 - 3\sin x}{\cos^2 x} dx = \int \left( \frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x} \right) dx \)
\( = \int (2\sec^2 x - 3\sec x \tan x) dx = 2\tan x - 3\sec x + C \)

Question. Write the value of \( \int \sec x (\sec x + \tan x) dx \) (Delhi 2011)
Answer: \( \int \sec x (\sec x + \tan x) dx = \int (\sec^2 x + \sec x \tan x) dx = \tan x + \sec x + C \)

Question. Write the value of \( \int (ax + b)^3 dx \) (AI 2011)
Answer: \( \int (ax + b)^3 dx = \frac{(ax + b)^4}{4a} + C \)

Question. Evaluate : \( \int \frac{x^3 - x^2 + x - 1}{x - 1} dx \) (Delhi 2011C)
Answer: Let \( I = \int \frac{x^3 - x^2 + x - 1}{x - 1} dx \)
\( = \int \frac{x^2(x - 1) + 1(x - 1)}{x - 1} dx = \int \frac{(x^2 + 1)(x - 1)}{x - 1} dx \)
\( = \int (x^2 + 1) dx = \frac{1}{3}x^3 + x + C \)

Question. Evaluate : \( \int \frac{x^3 - 1}{x^2} dx \) (Delhi 2010 C)
Answer: \( \int \frac{x^3 - 1}{x^2} dx = \int \left( \frac{x^3}{x^2} - \frac{1}{x^2} \right) dx \)
\( = \int \left( x - \frac{1}{x^2} \right) dx = \frac{x^2}{2} + \frac{1}{x} + C \)

Question. Write the value of \( \int \frac{1 - \sin x}{\cos^2 x} dx \) (AI 2010 C)
Answer: Refer to answer 7.

Question. Write the value of \( \int 2^x dx \) (AI 2010 C)
Answer: \( \int 2^x dx = \frac{2^x}{\log 2} + C \)

Question. Find : \( \int \frac{\sin^6 x}{\cos^8 x} dx \) (AI 2014C)
Answer: Let \( I = \int \frac{\sin^6 x}{\cos^8 x} dx = \int \frac{\sin^6 x}{\cos^6 x \cdot \cos^2 x} dx \)
\( = \int \tan^6 x \sec^2 x dx \)
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( \therefore I = \int t^6 dt = \frac{t^7}{7} + C = \frac{1}{7} \tan^7 x + C \)

Question. Write the value of : \( \int \frac{x + \cos 6x}{3x^2 + \sin 6x} dx \) (AI 2012C)
Answer: Let \( I = \int \frac{x + \cos 6x}{3x^2 + \sin 6x} dx \)
Put \( 3x^2 + \sin 6x = t \)
\( \Rightarrow (6x + 6\cos 6x)dx = dt \)
\( \Rightarrow (x + \cos 6x)dx = \frac{1}{6}dt \)
\( \therefore I = \int \frac{1}{6} \frac{dt}{t} = \frac{1}{6}\log|t| + C \)
\( = \frac{1}{6}\log(3x^2 + \sin 6x) + C \)

Question. Evaluate : \( \int \frac{(\log x)^2}{x} dx \) (AI 2011)
Answer: Let \( I = \int \frac{(\log x)^2}{x} dx \)
Put \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \therefore I = \int t^2 dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C \)

Question. Evaluate : \( \int \frac{e^{\tan^{-1} x}}{1 + x^2} dx \) (AI 2011)
Answer: Let \( I = \int \frac{e^{\tan^{-1} x}}{1 + x^2} dx \)
Put \( \tan^{-1} x = t \Rightarrow \frac{1}{1 + x^2} dx = dt \)
\( \therefore I = \int e^t dt = e^t + C = e^{\tan^{-1} x} + C \)

Question. Evaluate : \( \int \frac{2\cos x}{3\sin^2 x} dx \) (AI 2011C)
Answer: Let \( I = \int \frac{2\cos x}{3\sin^2 x} dx \)
Put \( \sin x = t \Rightarrow \cos x dx = dt \)
\( \therefore I = \frac{2}{3} \int \frac{dt}{t^2} = \frac{2}{3} \int t^{-2} dt = \frac{2}{3} \frac{t^{-1}}{-1} + C = -\frac{2}{3\sin x} + C \)

Question. Evaluate : \( \int \frac{\log x}{x} dx \) (Delhi 2010)
Answer: Let \( I = \int \frac{\log x}{x} dx \)
Put \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \dots I = \int t dt = \frac{t^2}{2} + C = \frac{(\log x)^2}{2} + C \)

Question. Evaluate : \( \int \sec^2(7 - 4x) dx \) (AI 2010)
Answer: Let \( I = \int \sec^2(7 - 4x) dx \)
Put \( 7 - 4x = t \Rightarrow dx = -\frac{1}{4}dt \)
\( \therefore I = \int \sec^2 t \left( -\frac{1}{4} \right) dt \Rightarrow I = -\frac{\tan t}{4} + C = -\frac{\tan(7 - 4x)}{4} + C \)

Question. Evaluate : \( \int \cos 4x \cos 3x \, dx \) (Delhi 2007)
Answer: Let \( I = \int \cos 4x \cos 3x \, dx \)
\( I = \frac{1}{2} \int 2\cos 4x \cos 3x \, dx \)
\( I = \frac{1}{2} \int [\cos(4x + 3x) + \cos(4x - 3x)] dx \)
\( I = \frac{1}{2} \int [\cos 7x + \cos x] dx \)
\( I = \frac{1}{2} \left[ \frac{\sin 7x}{7} + \sin x \right] + C \)
\( I = \frac{1}{14}\sin 7x + \frac{1}{2}\sin x + C \)

Question. Find \( \int \frac{(3\sin \theta - 2)\cos \theta}{5 - \cos^2 \theta - 4\sin \theta} d\theta \) (Delhi 2016, 2013C)
Answer: Let \( I = \int \frac{(3\sin \theta - 2)\cos \theta}{5 - \cos^2 \theta - 4\sin \theta} d\theta \)
\( = 3\int \frac{\sin \theta \cos \theta}{4 + \sin^2 \theta - 4\sin \theta} d\theta - 2\int \frac{\cos \theta}{4 + \sin^2 \theta - 4\sin \theta} d\theta \)
\( = 3I_1 - 2I_2 \) (say)
Now, \( I_1 = \int \frac{\sin \theta \cos \theta}{4 + \sin^2 \theta - 4\sin \theta} d\theta \)
Put \( \sin^2 \theta = t \Rightarrow 2\sin\theta\cos\theta d\theta = dt \)
\( \therefore I_1 = \frac{1}{2}\int \frac{dt}{4 + t - 4\sqrt{t}} = \frac{1}{2}\int \frac{dt}{(\sqrt{t} - 2)^2} \)
Put \( \sqrt{t} - 2 = u \Rightarrow \sqrt{t} = u + 2 \)
\( \Rightarrow \frac{1}{2\sqrt{t}} dt = du \Rightarrow dt = 2\sqrt{t}du \Rightarrow dt = 2(u + 2)du \)
\( \therefore I_1 = \int \frac{(u + 2)}{u^2} du = \int \frac{du}{u} + 2\int \frac{du}{u^2} \)
\( = \log|u| - \frac{2}{u} + C_1 = \log|\sqrt{t} - 2| - \frac{2}{\sqrt{t} - 2} + C_1 \)
\( = \log|\sin \theta - 2| - \frac{2}{\sin \theta - 2} + C_1 \)
Also, \( I_2 = \int \frac{\cos \theta}{4 + \sin^2 \theta - 4\sin \theta} d\theta \)
Put \( \sin \theta = m \Rightarrow \cos \theta d\theta = dm \)
\( \therefore I_2 = \int \frac{dm}{4 + m^2 - 4m} = \int \frac{dm}{(m - 2)^2} \)
\( = \frac{-1}{m-2} + C_2 = \frac{-1}{\sin \theta - 2} + C_2 \)
\( \therefore I = 3\log|\sin \theta - 2| - \frac{6}{\sin \theta - 2} - \frac{2}{\sin \theta - 2} + C \), where \( C = 3C_1 - 2C_2 \)
\( \Rightarrow I = 3\log|\sin \theta - 2| - \frac{4}{\sin \theta - 2} + C \)

Question. Evaluate : \( \int \frac{\sin(x - a)}{\sin(x + a)} dx \) (Foreign 2015, Delhi 2013)
Answer: Let \( I = \int \frac{\sin(x - a)}{\sin(x + a)} dx = \int \frac{\sin(x + a - 2a)}{\sin(x + a)} dx \)
\( = \int \frac{\sin(x + a)\cos 2a - \cos(x + a)\sin 2a}{\sin(x + a)} dx \)
\( \Rightarrow I = \cos 2a \int dx - \sin 2a \int \frac{\cos(x + a)}{\sin(x + a)} dx \)
Put \( \sin(x + a) = t \Rightarrow \cos(x + a)dx = dt \)
\( \Rightarrow I = \cos 2a \int dx - \sin 2a \int \frac{dt}{t} \)
\( = x \cos 2a - \sin 2a \log|\sin(x + a)| + C \)

Question. Evaluate : \( \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \) (Delhi 2014)
Answer: \( \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \frac{\sin^6 x}{\sin^2 x \cos^2 x} dx + \int \frac{\cos^6 x}{\sin^2 x \cos^2 x} dx \)
\( = \int \frac{\sin^4 x}{\cos^2 x} dx + \int \frac{\cos^4 x}{\sin^2 x} dx \)
\( = \int \frac{\sin^2 x(1 - \cos^2 x)}{\cos^2 x} dx + \int \frac{\cos^2 x(1 - \sin^2 x)}{\sin^2 x} dx \)
\( = \int [\tan^2 x - \sin^2 x] dx + \int [\cot^2 x - \cos^2 x] dx \)
\( = \int (\sec^2 x - 1)dx + \int (\operatorname{cosec}^2 x - 1)dx - \int \sin^2 x \, dx - \int (1 - \sin^2 x) dx \)
\( = \tan x - x + (-\cot x) - x - x + C = \tan x - \cot x - 3x + C \)

Question. Evaluate : \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \) (AI 2013)
Answer: Here \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \)
\( = \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx \)
\( = 2 \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} dx \)
\( = 2 \int \frac{(\cos x + \cos \alpha)(\cos x - \cos \alpha)}{\cos x - \cos \alpha} dx \)
\( = 2 \int (\cos x + \cos \alpha) dx = 2[\sin x + x \cos \alpha] + C \)

Question. Evaluate : \( \int \sin x \sin 2x \sin 3x \, dx \) (Delhi 2012)
Answer: \( \int \sin x \sin 2x \sin 3x \, dx \)
\( = \int \sin 3x \sin 2x \sin x \, dx \)
\( = \frac{1}{2} \int (\cos 2x - \cos 4x)\sin 2x \, dx \)
\( = \frac{1}{2} \int \sin 2x \cos 2x \, dx - \frac{1}{2} \int \cos 4x \sin 2x \, dx \)
\( = \frac{1}{4} \int \sin 4x \, dx - \frac{1}{4} \int (\sin 6x - \sin 2x) dx \)
\( = \frac{1}{4} \int \sin 4x \, dx - \frac{1}{4} \int \sin 6x \, dx + \frac{1}{4} \int \sin 2x \, dx \)
\( = \frac{1}{4} \left[ \frac{-\cos 4x}{4} - \frac{(-\cos 6x)}{6} + \frac{(-\cos 2x)}{2} \right] + C \)
\( = \frac{1}{4} \left[ \frac{\cos 6x}{6} - \frac{\cos 4x}{4} - \frac{\cos 2x}{2} \right] + C \)

Question. Write the value of \( \int \frac{dx}{x^2 + 16} \) (Delhi 2011)
Answer: \( \int \frac{dx}{x^2 + (4)^2} = \frac{1}{4}\tan^{-1}\frac{x}{4} + C \)

Question. Evaluate : \( \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} dx \)
Answer: Let \( I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} dx \)
Put \( x^{3/2} = t \Rightarrow \frac{3}{2}\sqrt{x} dx = dt \Rightarrow \sqrt{x} dx = \frac{2}{3}dt \)
\( \therefore I = \frac{2}{3}\int \frac{dt}{\sqrt{(a^{3/2})^2 - t^2}} = \frac{2}{3}\sin^{-1}\left( \frac{t}{a^{3/2}} \right) + C = \frac{2}{3}\sin^{-1}\left( \frac{x^{3/2}}{a^{3/2}} \right) + C \)