CBSE Class 12 Mathematics Integration Worksheet Set 06

CBSE Class 12 Mathematics Integration Worksheet (4). 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. \( I = \int \sin^2 x \, dx \)
Answer: \( I = \int \sin^2 x \, dx \)
\( = \frac{1}{2} \int 1 - \cos(2x) \, dx \)
\( = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right] + c \)

Question. \( I = \int \tan^2(2x) \, dx \)
Answer: \( I = \int \tan^2(2x) \, dx \)
\( = \int \sec^2(2x) - 1 \, dx \)
\( = \frac{\tan(2x)}{2} - x + c \)

Question. \( I = \int \cot^2(3x) \, dx \)
Answer: \( I = \int \cot^2(3x) \, dx \)
\( = \int \csc^2(3x) - 1 \, dx \)
\( = -\frac{1}{3} \cot(3x) - x + c \)

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

Question. \( I = \int \sin^3 x \, dx \)
Answer: \( I = \int \sin^3 x \, dx \)
\( = \frac{1}{4} \int 3\sin x - \sin(3x) \, dx \)
\( = \frac{1}{4} \left[ -3\cos x + \frac{\cos(3x)}{3} \right] + c \)

Question. \( I = \int \cos^3(2x) \, dx \)
Answer: \( I = \int \cos^3(2x) \, dx \)
\( = \frac{1}{4} \int 3\cos(2x) + \cos(6x) \, dx \)
\( = \frac{1}{4} \left[ \frac{3}{2}\sin(2x) + \frac{1}{6}\sin(6x) \right] + c \)

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

Question. \( I = \int \cot^3(3x) \, dx \)
Answer: \( I = \int \cot^3(3x) \, dx \)
\( = \int \cot(3x) \cdot \cot^2(3x) \, dx \)
\( = \int \cot(3x) \cdot (\csc^2(3x) - 1) \, dx \)
\( = \int \cot(3x) \cdot \csc^2(3x) \, dx - \int \cot(3x) \, dx \)
put \( \cot(3x) = t \)
\( \therefore -\csc^2(3x) \cdot 3 \, dx = dt \)
\( \csc^2(3x) \, dx = -\frac{dt}{3} \)
\( \therefore I = -\frac{1}{3}\int t \, dt - \frac{1}{3}\log|\sin(3x)| \)
\( = -\frac{1}{6}t^2 - \frac{1}{3}\log|\sin(3x)| + c \)
Replacing \( t \)
\( I = -\frac{1}{6}\cot^2(3x) - \frac{1}{3}\log|\sin(3x)| + c \)

Question. \( I = \int \sin^4 x \, dx \)
Answer: \( I = \int \sin^4 x \, dx \)
\( = \int (\sin^2 x)^2 \, dx \)
\( = \int \left(\frac{1 - \cos(2x)}{2}\right)^2 \, dx \)
\( = \frac{1}{4} \int 1 + \cos^2(2x) - 2\cos(2x) \, dx \)
\( = \frac{1}{4} \int 1 + \frac{1 + \cos(4x)}{2} - 2\cos(2x) \, dx \)
\( = \frac{1}{8} \int 2 + 1 + \cos(4x) - 4\cos(2x) \, dx \)
\( = \frac{1}{8} \int 3 + \cos(4x) - 4\cos(2x) \, dx \)
\( I = \frac{1}{8} \left[ 3x + \frac{\sin(4x)}{4} - 2\sin(2x) \right] + c \)

Question. \( I = \int \cos^4(2x) \, dx \)
Answer: \( I = \int \cos^4(2x) \, dx \)
\( = \int (\cos^2(2x))^2 \, dx \)
\( = \int \left(\frac{1 + \cos(4x)}{2}\right)^2 \, dx \)
\( = \frac{1}{4} \int 1 + \cos^2(4x) + 2\cos(4x) \, dx \)
\( = \frac{1}{4} \int 1 + \frac{1 + \cos(8x)}{2} + 2\cos(4x) \, dx \)
\( = \frac{1}{8} \int 3 + \cos(8x) + 4\cos(4x) \, dx \)
\( I = \frac{1}{8} \left[ 3x + \frac{\sin(8x)}{8} + \frac{4}{4}\sin(4x) \right] + c \)

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

Question. \( I = \int \cot^4(3x) \, dx \)
Answer: \( I = \int \cot^4(3x) \, dx \)
\( = \int \cot^2(3x) \cdot \cot^2(3x) \, dx \)
\( = \int \cot^2(3x) \cdot (\csc^2(3x) - 1) \, dx \)
\( = \int \cot^2(3x) \cdot \csc^2(3x) \, dx - \int \cot^2(3x) \, dx \)
put \( \cot(3x) = t \)
\( -3\csc^2(3x) \, dx = dt \)
\( \therefore \csc^2(3x) \, dx = -\frac{dt}{3} \)
\( I = -\frac{1}{3}\int t^2 \, dt - \int \csc^2(3x) - 1 \, dx \)
\( = -\frac{1}{3} \cdot \frac{t^3}{3} - \left( -\frac{\cot(3x)}{3} - x \right) + c \)
\( I = -\frac{1}{9}\cot^2(3x) - \frac{\cot(3x)}{3} + x + c \)

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

Sin x and Cos x in multiplication with same power :-

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

Question. \( I = \int \sin^4 x \cdot \cos^4 x \, dx \)
Answer: \( I = \int \sin^4 x \cdot \cos^4 x \, dx \)
\( = \int (\sin x \cdot \cos x)^4 \, dx \)
\( = \int \left( \frac{\sin(2x)}{2} \right)^4 \, dx \)
\( = \frac{1}{16} \int \sin^4(2x) \, dx \)
\( = \frac{1}{16} \int (\sin^2(2x))^2 \, dx \)
\( = \frac{1}{16} \int \left( \frac{1 - \cos(4x)}{2} \right)^2 \, dx \)
\( = \frac{1}{64} \int 1 + \cos^2(4x) - 2\cos(4x) \, dx \)
\( = \frac{1}{64} \int 1 + \frac{1 + \cos(8x)}{2} - 2\cos(4x) \, dx \)
\( = \frac{1}{128} \int 3 + \cos(8x) - 4\cos(4x) \, dx \)
\( I = \frac{1}{128} \left[ 3x + \frac{\sin(8x)}{8} - \sin(4x) \right] + c \)

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

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

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

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

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

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