CBSE Class 12 Mathematics Integration Worksheet Set 11

CBSE Class 12 Mathematics Integration Worksheet (9). CBSE issues sample papers every year for students for class 12 board exams. Students should solve the CBSE issued sample papers to understand the pattern of the question paper which will come in class 12 board exams this year. The sample papers have been provided with marking scheme. It’s always recommended to practice as many CBSE sample papers as possible before the board examinations. Sample papers should be always practiced in examination condition at home or school and the student should show the answers to teachers for checking or compare with the answers provided. Students can download the sample papers in pdf format free and score better marks in examinations. Refer to other links too for latest sample papers.

Integration (Indefinite Integrals)

Question. (a) \( I = \int \frac{2\sin 2x - \cos x}{6 - \cos^2 x - 4\sin x} \, dx \)      (b) \( I = \int \frac{1}{2e^{2x} + 3e^x + 1} \, dx \)
Answer:
(a) \( I = \int \frac{2\sin 2x - \cos x}{6 - \cos^2 x - 4\sin x} \, dx \)
\( = \int \frac{4\sin x \cos x - \cos x}{6 - \cos^2 x - 4\sin x} \, dx \)
\( = \int \frac{(4\sin x - 1)\cos x}{6 - (1 - \sin^2 x) - 4\sin x} \, dx \)
put \( \sin x = t \)
\( \cos x \, dx = dt \)
\( I = \int \frac{(4t - 1)}{6 - (1 - t^2) - 4t} \, dt \)
\( = \int \frac{4t - 1}{t^2 - 4t + 5} \, dt \)
Proceed Yourself
\( 2\log|\sin^2 x - 4\sin x + 5| + 7\tan^{-1}(\sin x - 2) + c \) ans.

(b) \( I = \int \frac{1}{2e^{2x} + 3e^x + 1} \, dx \)
\( = \int \frac{1}{e^{2x}\left(2 + 3e^{-x} + e^{-2x}\right)} \, dx \)
L.C.M
\( = \int \frac{e^{-2x}}{2 + 3e^{-x} + e^{-2x}} \, dx \)
\( = \int \frac{e^{-x} \cdot e^{-x}}{2 + 3e^{-x} + e^{-2x}} \, dx \)
put \( e^{-x} = t \)
\( e^{-x} \, dx = -dt \)
\( \therefore I = -\int \frac{t}{2 + 3t + t^2} \, dt \)
\( = -2\int \frac{2t+3-3}{t^2+3t+2} \, dt \) (to make \( 2t + 3 \))
\( = -2\int \frac{2t+3}{t^2+3t+2} \, dt + 6\int \frac{1}{t^2+3t+2} \, dt \)
put \( t^2 + 3t + 2 = z \)
\( (2t+3) \, dt = dz \)
\( \therefore I = -2\int \frac{dz}{z} + 6\int \frac{1}{\left(t+\frac{3}{2}\right)^2 - \frac{9}{4} + 2} \, dt \)
\( = -2\log|z| + 6\int \frac{1}{\left(t+\frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \, dt \)
\( = -2\log|t^2+3t+2| + 6 \times \frac{1}{2 \times \frac{1}{2}} \log\left| \frac{t+\frac{3}{2}-\frac{1}{2}}{t+\frac{3}{2}+\frac{1}{2}} \right| + c \)
\( = -2\log|t^2+3t+2| + 6\log\left| \frac{2t+2}{2t+4} \right| + c \)
replacing \( t \) by \( e^{-x} \)
\( = I = -2\log|e^{-2x} + 3e^{-x} + 2| + 6\log\left| \frac{e^{-x}+1}{e^{-x}+2} \right| + c \) ans.

Question. (a) \( I = \int \frac{3x-1}{\sqrt{1-x-x^2}} \, dx \)      (b) \( I = \int \sqrt{\frac{1+x}{x}} \, dx \)
Answer:
(a) \( I = \int \frac{3x-1}{\sqrt{1-x-x^2}} \, dx \)
\( = 3\int \frac{x-\frac{1}{3}}{\sqrt{1-x-x^2}} \, dx \) (to make \( (-2x-1) \))
\( = -\frac{3}{2}\int \frac{-2x+\frac{2}{3}}{\sqrt{1-x-x^2}} \, dx \)
\( = -\frac{3}{2}\int \frac{-2x+\frac{2}{3}-1+1}{\sqrt{1-x-x^2}} \, dx \)
\( = -\frac{3}{2}\int \frac{(-2x-1)+\frac{5}{3}}{\sqrt{1-x-x^2}} \, dx \)
\( = -\frac{3}{2}\int \frac{(-2x-1)}{\sqrt{1-x-x^2}} \, dx - \frac{5}{2}\int \frac{1}{\sqrt{1-x-x^2}} \, dx \)
put \( 1-x-x^2 = t \)
\( (-2x-1) \, dx = dt \)
\( \therefore I = -\frac{3}{2}\int \frac{dt}{\sqrt{t}} - \frac{5}{2}\int \frac{1}{\sqrt{-[x^2+x-1]}} \, dx \)
\( = -\frac{3}{2} \times 2\sqrt{t} - \frac{5}{2}\int \frac{1}{\sqrt{-\left[\left(x+\frac{1}{2}\right)^2 - \frac{1}{4} - 1\right]}} \, dx \)
\( = -3\sqrt{t} - \frac{5}{2}\int \frac{1}{\sqrt{-\left[\left(x+\frac{1}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^2\right]}} \, dx \)
\( = -3\sqrt{1-x-x^2} - \frac{5}{2}\int \frac{1}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x+\frac{1}{2}\right)^2}} \, dx \)
\( = -3\sqrt{1-x-x^2} - \frac{5}{2}\sin^{-1}\left(\frac{2x+1}{\sqrt{5}}\right) + c \) ans.

(b) \( I = \int \sqrt{\frac{1+x}{x}} \, dx \)
\( = \int \sqrt{\frac{1+x}{x} \times \frac{1+x}{1+x}} \, dx \)
\( = \int \frac{1+x}{\sqrt{x+x^2}} \, dx \) (make \( 2x+1 \))
Proceed yourself
\( \sqrt{x^2+x} + \frac{1}{2}\log\left|\left(x+\frac{1}{2}\right) + \sqrt{x^2+x}\right| + c \) ans.

Question. (a) \( I = \int \sqrt{\frac{a-x}{a+x}} \, dx \)      (b) \( I = \int \sqrt{\frac{1-x}{1+x}} \, dx \)
Answer:
(a) \( I = \int \sqrt{\frac{a-x}{a+x}} \, dx \)
\( = \int \sqrt{\frac{a-x}{a+x} \times \frac{a-x}{a-x}} \, dx \)
\( = \int \frac{a-x}{\sqrt{a^2-x^2}} \, dx \)
Separate
\( = a\int \frac{1}{\sqrt{a^2-x^2}} \, dx - \int \frac{x}{\sqrt{a^2-x^2}} \, dx \)
put \( a^2-x^2 = t \)
\( -2x \, dx = dt \)
\( x \, dx = -\frac{dt}{2} \)
\( = a\sin^{-1}\left(\frac{x}{a}\right) + \frac{1}{2}\int \frac{dt}{\sqrt{t}} \)
\( = a\sin^{-1}\left(\frac{x}{a}\right) + \frac{1}{2} \times 2\sqrt{t} + c \)
\( = a\sin^{-1}\left(\frac{x}{a}\right) + \sqrt{a^2-x^2} + c \) ans.

Type: Divide by \(\cos^2 x\)

Question. (a) \( I = \int \frac{1}{1+3\sin^2 x+8\cos^2 x} \, dx \)      (b) \( I = \int \frac{1}{3+\sin(2x)} \, dx \)
Answer:
(a) \( I = \int \frac{1}{1+3\sin^2 x+8\cos^2 x} \, dx \)
Divide by \(\cos^2 x\)
\( I = \int \frac{\sec^2 x}{\sec^2 x + 3\tan^2 x + 8} \, dx \)
\( = \int \frac{\sec^2 x \, dx}{(1+\tan^2 x)+3\tan^2 x+8} \)
put \( \tan x = t \)
\( \sec^2 x \, dx = dt \)
\( I = \int \frac{dt}{(1+t^2)+3t^2+8} \)
\( = \int \frac{dt}{4t^2+9} \)
\( = \frac{1}{4}\int \frac{1}{t^2+\left(\frac{3}{2}\right)^2} \, dt \)
\( = \frac{1}{4} \times \frac{2}{3}\tan^{-1}\left(\frac{2t}{3}\right) + c \)
\( I = \frac{1}{6}\tan^{-1}\left(\frac{2\tan x}{3}\right) + c \) ans.

(b) \( I = \int \frac{1}{3+\sin(2x)} \, dx \)
\( = \int \frac{1}{3+2\sin x \cos x} \, dx \)
Divide N & D by \(\cos^2 x\)
\( = \int \frac{\sec^2 x}{3\sec^2 x + 2\tan x} \, dx \)
\( = \int \frac{\sec^2 x}{3(1+\tan^2 x) + 2\tan x} \, dx \)
put \( \tan x = t \)
\( \sec^2 x \, dx = dt \)
\( I = \int \frac{dt}{3(1+t^2) + 2t} \)
\( = \int \frac{dt}{3t^2+2t+3} \)
perfect square:
Proceed Your self
\( \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{3\tan x+1}{2\sqrt{2}}\right) + c \) ans.

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

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

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

Type: Single \(\sin x\), \(\cos x\)

Question. (a) \( I = \int \frac{1}{1 + \sin x + \cos x} \, dx \)      (b) \( I = \int \frac{1}{\sin x - \sqrt{3}\cos x} \, dx \)
Answer:
(a) \( I = \int \frac{1}{1 + \sin x + \cos x} \, dx \)
\( = \int \frac{1}{1 + \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} + \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} \, dx \)
\( = \int \frac{1 + \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right) + 2\tan\left(\frac{x}{2}\right) + 1 - \tan^2\left(\frac{x}{2}\right)} \, dx \)
\( = \int \frac{\sec^2\left(\frac{x}{2}\right)}{2 + 2\tan\left(\frac{x}{2}\right)} \, dx \)
put \( \tan\frac{x}{2} = t \)
\( \therefore \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \, dx = dt \)
\( \sec^2\left(\frac{x}{2}\right) \, dx = 2dt \)
\( \therefore I = 2\int \frac{dt}{2+2t} \)
\( = \frac{2}{2}\log|2+2t| + c \)
\( = \log\left|2 + 2\tan\left(\frac{x}{2}\right)\right| + c \) ans.

(b) \( I = \int \frac{1}{\sin x - \sqrt{3}\cos x} \, dx \)
\( = \int \frac{1}{\frac{2\tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)} - \frac{\sqrt{3}\left(1-\tan^2\frac{x}{2}\right)}{1+\tan^2\frac{x}{2}}} \, dx \)
\( = \int \frac{1+\tan^2\left(\frac{x}{2}\right)}{2\tan\left(\frac{x}{2}\right) - \sqrt{3} + \sqrt{3}\tan^2\left(\frac{x}{2}\right)} \, dx \)
\( = \int \frac{\sec^2\left(\frac{x}{2}\right)}{2\tan\left(\frac{x}{2}\right) - \sqrt{3} + \sqrt{3}\tan^2\left(\frac{x}{2}\right)} \, dx \)
put \( \tan\left(\frac{x}{2}\right) = t \)
\( \sec^2\left(\frac{x}{2}\right) \, dx = 2 \, dt \)
\( \therefore I = 2\int \frac{dt}{\sqrt{3}t^2 + 2t - \sqrt{3}} \)
\( I = \frac{2}{\sqrt{3}}\int \frac{dt}{t^2 + \frac{2}{\sqrt{3}}t - 1} \)
\( = \frac{2}{\sqrt{3}}\int \frac{dt}{\left(t+\frac{1}{\sqrt{3}}\right)^2 - \frac{1}{3} - 1} \)
\( = \frac{2}{\sqrt{3}}\int \frac{dt}{\left(t+\frac{1}{\sqrt{3}}\right)^2 - \left(\frac{2}{\sqrt{3}}\right)^2} \)
\( = \frac{2}{\sqrt{3}} \times \frac{1}{2 \times \frac{2}{\sqrt{3}}} \log\left| \frac{t + \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}} \right| + c \)
\( = \frac{1}{2}\log\left| \frac{\sqrt{3}t - 1}{\sqrt{3}t + 3} \right| + c \)
\( = \frac{1}{2}\log\left| \frac{\sqrt{3}\tan\left(\frac{x}{2}\right) - 1}{\sqrt{3}\tan\left(\frac{x}{2}\right) + 3} \right| + c \) ans.

Type: \(\int \frac{a\sin x + b\cos x}{c\sin x + d\cos x} \, dx\)

Question. \( I = \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx \)
Answer:
\( I = \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx \)
let \( 3\sin x + 2\cos x = A\left[\frac{d}{dx}(3\cos x + 2\sin x)\right] + B[3\cos x + 2\sin x] \)
\( 3\sin x + 2\cos x = A(-3\sin x + 2\cos x) + B(3\cos x + 2\sin x) \)
equating coefficients of \(\sin x\) and \(\cos x\) on both sides
\( 3 = -3A + 2B \)
\( 2 = 2A + 3B \)
solving these two equations, we get
\( A = -\frac{5}{13} \) and \( B = \frac{12}{13} \)
\( \therefore I = \int \frac{-\frac{5}{13}(-3\sin x + 2\cos x) + \frac{12}{13}(3\cos x + 2\sin x)}{3\cos x + 2\sin x} \, dx + c \)
Separate
\( I = -\frac{5}{13}\int \frac{-3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx + \frac{12}{13}\int dx \)
put \( 3\cos x + 2\sin x = t \)
\( (-3\sin x + 2\cos x) \, dx = dt \)
\( \therefore I = -\frac{5}{13}\int \frac{dt}{t} + \frac{12}{13}x \)
\( I = -\frac{5}{13}\log|3\cos x + 2\sin x| + \frac{12}{13}x + c \) ans.

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