CBSE Class 12 Mathematics Integration Worksheet Set 02

CBSE Class 12 Mathematics Integration Worksheet (1). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Integration (Indefinite Integrals)

Question. \( I = \int x\cos^3(x^2)\sin(x^2) \, dx \)
Answer: \( I = \int x\cos^3(x^2)\sin(x^2) \, dx \)
put \( x^2 = t \)
\( 2x \, dx = dt \Rightarrow x \, dx = \frac{1}{2} \, dt \)
\( \therefore I = \frac{1}{2} \int \cos^3 t \cdot \sin t \, dt \)
put \( \cos t = z \)
\( \sin t \, dt = -dz \)
\( \dots I = -\frac{1}{2} \int z^3 \, dz \)
\( = -\frac{1}{2} \frac{z^4}{4} + c \)
replacing 'z'
\( = -\frac{1}{8} \cos^4 t + c \)
replacing t
\( = -\frac{1}{8} \cos^4(x^2) + c \)

Question. \( I = \int \frac{1}{\sin^3 x \cdot \cos x} \, dx \)
Answer: \( I = \int \frac{1}{\sin^3 x \cdot \cos x} \, dx \)
Divide N & D by \( \cos^4 x \)
\( I = \int \frac{\sec^4 x}{\tan^3 x} \, dx \)
\( = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^3 x} \, dx \)
\( = \int \frac{(1+\tan^2 x)\cdot\sec^2 x}{\tan^3 x} \, dx \)
put \( \tan x = t \)
\( \therefore \sec^2 x \, dx = dt \)
\( I = \int \frac{(1+t^2)}{t^3} \, dt \)
\( = \int \left(\frac{1}{t^3} + \frac{1}{t}\right) \, dt \)
\( = \log |t| - \frac{1}{2t^2} + c \)
\( I = \log |\tan x| - \frac{1}{2}\cot^2 x + c \)

Question. \( I = \int \frac{1}{\sin^4 x \cdot \cos^2 x} \, dx \)
Answer: \( I = \int \frac{1}{\sin^4 x \cdot \cos^2 x} \, dx \)
Divide N & D by \( \cos^6 x \)
\( = \int \frac{\sec^6 x}{\tan^4 x} \, dx \)
\( = \int \frac{\sec^4 x \cdot \sec^2 x}{\tan^4 x} \, dx \)
\( = \int \frac{(1+\tan^2 x)^2 \cdot \sec^2 x}{\tan^4 x} \, dx \)
put \( \tan x = t \)
\( \sec^2 x \, dx = dt \)
\( \therefore I = \int \frac{(1+t^2)^2}{t^4} \, dt \)
\( = \int \frac{1+t^4+2t^2}{t^4} \, dt \)
\( = \int \left( \frac{1}{t^4} + 1 + \frac{2}{t^2} \right) \, dt \)
\( = -\frac{1}{3t^3} + t - \frac{2}{t} + c \)
\( \therefore I = -\frac{1}{3}\cot^3 x + \tan x - 2\cot x + c \)

Question. \( I = \int \frac{\cos^9 x}{\sin x} \, dx \)
Answer: \( I = \int \frac{\cos^9 x}{\sin x} \, dx \)
\( = \int \frac{\cos^8 x \cdot \cos x}{\sin x} \, dx \)
\( = \int \frac{(\cos^2 x)^4 \cdot \cos x}{\sin x} \, dx \)
\( = \int \frac{(1-\sin^2 x)^4 \cos x}{\sin x} \, dx \)
put \( \sin x = t \)
\( \therefore \cos x \, dx = dt \)
\( = \int \frac{(1-t^2)^4}{t} \, dt \)
\( = \int \frac{(1+t^4-2t^2)^2}{t} \, dt \)
\( = \int \frac{1+t^8+4t^4+2t^4-4t^6-4t^2}{t} \, dt \)
\( = \int \frac{t^8-4t^6+6t^4-4t^2+1}{t} \, dt \)
\( = \int \left( t^7 - 4t^5 + 6t^3 - 4t + \frac{1}{t} \right) \, dt \)
\( = \frac{t^8}{8} - \frac{4t^6}{6} + \frac{6t^4}{4} - \frac{4t^2}{2} + \log |t| + c \)
\( = \frac{\sin^8 x}{8} - \frac{2}{3}\sin^6 x + \frac{3}{2}\sin^4 x - 2\sin^2 x + \log |\sin x| + c \)

Sin x and Cos x in multiplication with different Angles :-

Question. \( I = \int \sin(3x)\cos(2x) \, dx \)
Answer: \( I = \int \sin(3x)\cos(2x) \, dx \)
\( = \frac{1}{2} \int 2\sin(3x)\cos(2x) \, dx \)
\( = \frac{1}{2} \int (\sin(5x) + \sin x) \, dx \)
\( I = \frac{1}{2} \left[ -\frac{\cos(5x)}{5} - \cos x \right] + c \)

Question. \( I = \int \sin(2x)\sin(4x)\sin(6x) \, dx \)
Answer: \( I = \int \sin(2x)\sin(4x)\sin(6x) \, dx \)
\( = \frac{1}{2} \int [2\sin(2x)\sin(4x)]\sin(6x) \, dx \)
\( = \frac{1}{2} \int [\cos(2x) - \cos(6x)]\sin(6x) \, dx \)
\( = \frac{1}{2} \int (\sin(6x)\cos(2x) - \sin(6x)\cos(6x)) \, dx \)
\( = \frac{1}{4} \int (2\sin(6x)\cos(2x) - 2\sin(6x)\cos(6x)) \, dx \)
\( = \frac{1}{4} \int (\sin(8x) + \sin(4x) - \sin(12x)) \, dx \)
\( = \frac{1}{4} \left[ -\frac{\cos(8x)}{8} - \frac{\cos(4x)}{4} + \frac{\cos(12x)}{12} \right] + c \)

Question. \( I = \int \frac{\sin(4x)}{\sin x} \, dx \)
Answer: \( I = \int \frac{\sin(4x)}{\sin x} \, dx \)
\( = 2 \int \frac{\sin(2x)\cos(2x)}{\sin x} \, dx \)
\( = 4 \int \frac{\sin x \cos x \cos(2x)}{\sin x} \, dx \)
\( = 2 \int 2\cos x \cos(2x) \, dx \)
\( = 2 \int (\cos(3x) + \cos x) \, dx \) ......{\( 2\cos A\cos B = \cos(A+B) + \cos(A-B) \)}
\( = 2 \left[ \frac{\sin(3x)}{3} + \sin x \right] + c \)

Question. \( I = \int \tan x \cdot \sec^4 x \, dx \)
Answer: \( I = \int \tan x \cdot \sec^4 x \, dx \)
\( = \int \tan x \cdot (1 + \tan^2 x)\cdot\sec^2 x \, dx \)
put \( \tan x = t \)
\( \therefore \sec^2 x \, dx = dt \)
\( I = \int t(1+t^2) \, dt \)
\( = \int (t+t^3) \, dt \)
\( = \frac{t^2}{2} + \frac{t^4}{4} + c \)
\( I = \frac{\tan^2 x}{2} + \frac{\tan^4 x}{4} + c \)

Question. \( I = \int \tan^3 x \cdot \sec^3 x \, dx \)
Answer: \( I = \int \tan^3 x \cdot \sec^3 x \, dx \)
\( = \int \tan^2 x \cdot \sec^2 x \cdot \tan x \sec x \, dx \)
\( = \int (\sec^2 x - 1)\cdot\sec^2 x \cdot \tan x \sec x \, dx \)
put \( \sec x = t \)
\( \sec x \tan x \, dx = dt \)
\( I = \int (t^2-1)t^2 \, dt \)
\( = \int (t^4 - t^2) \, dt \)
\( = \frac{t^5}{5} - \frac{t^3}{3} + c \)
\( I = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + c \)

Question. \( I = \int \sec^n x \cdot \tan x \, dx \)
Answer: \( I = \int \sec^n x \cdot \tan x \, dx \)
\( = \int \sec^{n-1} x \cdot \tan x \sec x \, dx \)
put \( \sec x = t \)
\( \sec x \tan x \, dx = dt \)
\( I = \int t^{n-1} \, dt \)
\( = \frac{t^n}{n} + c \)
\( I = \frac{\sec^n x}{n} + c \)

QNS Based On Sin(A \(\pm\) B) and Cos(A \(\pm\) B) :-

Question. \( I = \int \frac{\sin(x-a)}{\sin x} \, dx \)
Answer: \( I = \int \frac{\sin(x-a)}{\sin x} \, dx \)
\( = \int \frac{\sin x \cos a - \cos x \sin a}{\sin x} \, dx \)
Separate
\( = \int \cos a - \cot x \cdot \sin a \, dx \)
\( I = x \cos a - \log |\sin x| \cdot \sin a + c \)

Question. \( I = \int \frac{\sin x}{\sin(x+a)} \, dx \)
Answer: \( I = \int \frac{\sin x}{\sin(x+a)} \, dx \)
\( = \int \frac{\sin(x+a-a)}{\sin(x+a)} \, dx \)
\( = \int \frac{\sin(x+a)\cos a - \cos(x+a)\sin a}{\sin(x+a)} \, dx \)
\( = \int \cos a - \cot(x+a)\sin a \, dx \)
\( I = x \cos a - \log |\sin(x+a)| \cdot \sin a + c \)

Question. \( I = \int \frac{\sin(x+a)}{\sin(x+b)} \, dx \)
Answer: \( I = \int \frac{\sin(x+a)}{\sin(x+b)} \, dx \)
\( = \int \frac{\sin(x+a+b-b)}{\sin(x+b)} \, dx \)
\( = \int \frac{\sin[(x+b)+(a-b)]}{\sin(x+b)} \, dx \)
\( = \int \frac{\sin(x+b)\cos(a-b) + \cos(x+b)\sin(a-b)}{\sin(x+b)} \, dx \)
\( = \int \cos(a-b) + \cot(x+b)\sin(a-b) \, dx \)
\( I = x \cos(a-b) + \log |\sin(x+b)| \cdot \sin(a-b) + c \)

Question. \( I = \int \frac{\sin(2x)}{\sin(5x)\sin(3x)} \, dx \)
Answer: \( I = \int \frac{\sin(2x)}{\sin(5x)\sin(3x)} \, dx \)
\( = \int \frac{\sin(5x-3x)}{\sin(5x)\sin(3x)} \, dx \)
\( = \int \frac{\sin(5x)\cos(3x) - \cos(5x)\sin(3x)}{\sin(5x)\sin(3x)} \, dx \)
Separate
\( I = \int (\cot(3x) - \cot(5x)) \, dx \)
\( = \frac{1}{3}\log |\sin(3x)| - \frac{1}{5}\log |\sin(5x)| + c \)

Question. \( I = \int \frac{1}{\sin(x-a)\sin(x-b)} \, dx \)
Answer: \( I = \int \frac{1}{\sin(x-a)\sin(x-b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\sin(x-a)\sin(x-b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b+x-x)}{\sin(x-a)\sin(x-b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin[(x-b)-(x-a)]}{\sin(x-a)\sin(x-b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-a)\sin(x-b)}{\sin(x-a)\sin(x-b)} \, dx \)
Separate
\( = \frac{1}{\sin(a-b)} \int (\cot(x-a) - \cot(x-b)) \, dx \)
\( = \frac{1}{\sin(a-b)} [\log |\sin(x-a)| - \log |\sin(x-b)|] + c \)
\( I = \frac{1}{\sin(a-b)} \log \left|\frac{\sin(x-a)}{\sin(x-b)}\right| + c \)

Question. \( I = \int \frac{1}{\cos(x+a)\cos(x+b)} \, dx \)
Answer: \( I = \int \frac{1}{\cos(x+a)\cos(x+b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x+a)\cos(x+b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b+x-x)}{\cos(x+a)\cos(x+b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin[(x+a)-(x+b)]}{\cos(x+a)\cos(x+b)} \, dx \)
\( = \frac{1}{\sin(a-b)} \int \frac{\sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)} \, dx \)
Separate:
\( = \frac{1}{\sin(a-b)} \int (\tan(x+a) - \tan(x+b)) \, dx \)
\( = \frac{1}{\sin(a-b)} [\log |\sec(x+a)| - \log |\sec(x+b)|] + c \)
\( I = \frac{1}{\sin(a-b)} \log \left|\frac{\sec(x+a)}{\sec(x+b)}\right| + c \)

Question. \( I = \int \frac{1}{\sin(x-a)\cos(x-b)} \, dx \)
Answer: \( I = \int \frac{1}{\sin(x-a)\cos(x-b)} \, dx \)
\( = \frac{1}{\cos(a-b)} \int \frac{\cos(a-b)}{\sin(x-a)\cos(x-b)} \, dx \)
\( = \frac{1}{\cos(a-b)} \int \frac{\cos(a-b+x-x)}{\sin(x-a)\cos(x-b)} \, dx \)
\( = \frac{1}{\cos(a-b)} \int \frac{\cos[(x-b)-(x-a)]}{\sin(x-a)\cos(x-b)} \, dx \)
\( = \frac{1}{\cos(a-b)} \int \frac{\cos(x-b)\cos(x-a) + \sin(x-b)\sin(x-a)}{\sin(x-a)\cos(x-b)} \, dx \)
Separate
\( = \frac{1}{\cos(a-b)} \int (\cot(x-a) + \tan(x-b)) \, dx \)
\( I = \frac{1}{\cos(a-b)} [\log |\sin(x-a)| + \log |\sec(x-b)|] + c \)