CBSE Class 12 Mathematics Integration Worksheet Set 10

CBSE Class 12 Mathematics Integration Worksheet (8). 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. (i) \( I = \int \frac{x^4+1}{x^2+1} \, dx \)      (ii) \( I = \int \frac{x^3+x}{\sqrt{x^4+1}} \, dx \)
Answer:
(i) \( I = \int \frac{x^4+1}{x^2+1} \, dx \)
Degree of \( N^r > \text{degree of } D^r \)
\( = \int Q + \frac{R}{D} \, dx \)
\( = \int \left( x^2 - 1 + \frac{2}{x^2+1} \right) \, dx \)
\( I = \frac{x^3}{3} - x + 2\tan^{-1}x + c \)

Polynomial Division:
\[ \begin{array}{rll} & x^2 - 1 & \text{(Quotient)} \\ x^2 + 1 \!\!\!\! & \overline{) \ x^4 + 1} \\ & \underline{-(x^4 + x^2)} \\ & \ \ \ \ \ \ \ -x^2 + 1 \\ & \underline{\ \ \ \ \ -(-x^2 - 1)} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 & \text{(Remainder)} \end{array} \]

(ii) \( I = \int \frac{x^3+x}{\sqrt{x^4+1}} \, dx \)
\( = \int \frac{x^3}{x^4+1} \, dx + \int \frac{x}{x^4+1} \, dx \)
put \( x^4 + 1 = t \) in (I) and put \( x^2 = z \) in (II)
\( 4x^3 \, dx = dt \implies x^3 \, dx = \frac{dt}{4} \)
\( 2x \, dx = dz \implies x \, dx = \frac{dz}{2} \)
\( \therefore I = \frac{1}{4}\int \frac{dt}{t} + \frac{1}{2}\int \frac{dz}{z^2+1} \)
\( = \frac{1}{4}\log|t| + \frac{1}{2}\tan^{-1}(z) + c \)
\( = \frac{1}{4}\log|x^4+1| + \frac{1}{2}\tan^{-1}(x^2) + c \) ans.

Type: \(\int \frac{1}{\text{Quadratic}} \, dx\) — Make Perfect Square

Question. (i) \( I = \int \frac{1}{2x^2+x-1} \, dx \)      (ii) \( I = \int \frac{1}{3+2x-x^2} \, dx \)
Answer:
(i) \( I = \int \frac{1}{2x^2+x-1} \, dx \)
Make perfect square
\( = \frac{1}{2}\int \frac{1}{x^2+\frac{x}{2}-\frac{1}{2}} \, dx \)
\( = \frac{1}{2}\int \frac{1}{\left(x+\frac{1}{4}\right)^2-\frac{1}{16}-\frac{1}{2}} \, dx \)
\( = \frac{1}{2}\int \frac{1}{\left(x+\frac{1}{4}\right)^2-\frac{9}{16}} \, dx \)
\( = \frac{1}{2}\int \frac{1}{\left(x+\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2} \, dx \)
\( = \frac{1}{2} \times \frac{1}{2 \times \frac{3}{4}}\log\left|\frac{x+\frac{1}{4}-\frac{3}{4}}{x+\frac{1}{4}+\frac{3}{4}}\right| + c \qquad \dots\dots\dots \int \frac{1}{x^2-a^2} \, dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| \)
\( = \frac{1}{3}\log\left|\frac{4x-2}{4x+4}\right| + c \)
\( I = \frac{1}{3}\log\left|\frac{2x-1}{2x+2}\right| + c \) ans.

(ii) \( I = \int \frac{1}{3+2x-x^2} \, dx \)
Make a perfect square
\( = -\int \frac{1}{x^2-2x-3} \, dx \)
\( = -\int \frac{1}{(x-1)^2-1-3} \, dx \)
\( = -\int \frac{1}{(x-1)^2-(2)^2} \, dx \)
\( = \int \frac{1}{(2)^2-(x-1)^2} \, dx \)
\( = \frac{1}{2 \times 2}\log\left|\frac{2+x-1}{2-x+1}\right| + c \qquad \dots\dots\dots \int \frac{1}{a^2-x^2} \, dx = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| \)
\( = I = \frac{1}{4}\log\left|\frac{1+x}{3-x}\right| + c \) ans.

Question. (a) \( I = \int \frac{1}{\sqrt{(x-1)(x-2)}} \, dx \)      (b) \( I = \int \frac{1}{\sqrt{7-3x-2x^2}} \, dx \)      (c) \( I = \int \frac{1}{\sqrt{(x-a)(x-b)}} \, dx \)
Answer:
(a) \( I = \int \frac{1}{\sqrt{(x-1)(x-2)}} \, dx \)
\( I = \int \frac{1}{\sqrt{x^2-3x+2}} \, dx \)
\( = \int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+2}} \, dx \)
\( = \int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} \, dx \)
\( = \log\left| \left(x-\frac{3}{2}\right) + \sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} \right| + c \)
\( = I = \log\left| \left(x-\frac{3}{2}\right) + \sqrt{x^2-3x+2} \right| + c \) ans.

(b) \( I = \int \frac{1}{\sqrt{7-3x-2x^2}} \, dx \)
\( = \int \frac{1}{\sqrt{-2\left(x^2+\frac{3}{2}x-\frac{7}{2}\right)}} \, dx \)
\( = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{-\left[\left(x+\frac{3}{4}\right)^2-\frac{9}{16}-\frac{7}{2}\right]}} \, dx \)
\( = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{-\left[\left(x+\frac{3}{4}\right)^2-\left(\frac{\sqrt{65}}{4}\right)^2\right]}} \, dx \)
\( = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^2-\left(x+\frac{3}{4}\right)^2}} \, dx \)
\( = \frac{1}{\sqrt{2}}\sin^{-1}\left( \frac{x+\frac{3}{4}}{\frac{\sqrt{65}}{4}} \right) + c \qquad \dots\dots\dots \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) \)
\( = \frac{1}{\sqrt{2}}\sin^{-1}\left( \frac{4x+3}{\sqrt{65}} \right) + c \) ans.

(c) \( I = \int \frac{1}{\sqrt{(x-a)(x-b)}} \, dx \)
\( = \int \frac{1}{\sqrt{x^2-ax-bx+ab}} \, dx \)
\( = \int \frac{1}{\sqrt{x^2-(a+b)x+ab}} \, dx \)
\( = \int \frac{1}{\sqrt{\left\{x-\frac{(a+b)}{2}\right\}^2-\left(\frac{a+b}{2}\right)^2+ab}} \, dx \)
\( = \int \frac{1}{\sqrt{\left\{x-\frac{(a+b)}{2}\right\}^2-\left\{\frac{(a+b)^2}{4}-ab\right\}}} \, dx \)
\( = \int \frac{1}{\sqrt{\left\{x-\frac{(a+b)}{2}\right\}^2-\left\{\frac{a^2+b^2+2ab-4ab}{4}\right\}}} \, dx \)
\( = \int \frac{1}{\sqrt{\left\{x-\frac{(a+b)}{2}\right\}^2-\left\{\frac{(a-b)}{2}\right\}^2}} \, dx \)
\( = \log\left| \left(x-\frac{(a+b)}{2}\right) + \sqrt{\left\{x-\frac{a+b}{2}\right\}^2-\left(\frac{a-b}{2}\right)^2} \right| + c \)
\( = \log\left| \left(x-\frac{(a+b)}{2}\right) + \sqrt{(x-a)(x-b)} \right| + c \) ans.

Type: Separate \(\int \frac{\text{Linear}}{\text{Special Integral}} \, dx\)

Question. (a) \( I = \int \frac{3x-1}{x^2+4} \, dx \)      (b) \( I = \int \frac{3-2x}{\sqrt{x^2+4}} \, dx \)
Answer:
(a) \( I = \int \frac{3x-1}{x^2+4} \, dx \)
Separate
\( = 3\int \frac{x}{x^2+4} \, dx - \int \frac{1}{x^2+4} \, dx \)
put \( x^2 + 4 = t \) in (I)
\( 2x \, dx = dt \)
\( x \, dx = \frac{dt}{2} \)
\( \therefore I = 3 \times \frac{1}{2}\int \frac{dt}{t} - \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \)
\( = \frac{3}{2}\log|t| - \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \)
\( I = \frac{3}{2}\log|x^2+4| - \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \) ans.

(b) \( I = \int \frac{3-2x}{\sqrt{x^2+4}} \, dx \)
\( = 3\int \frac{1}{\sqrt{x^2+4}} \, dx - 2\int \frac{x}{\sqrt{x^2+4}} \, dx \)
put \( x^2+4 = t \) in (II)
\( 2x \, dx = dt \)
\( x \, dx = \frac{dt}{2} \)
\( I = 3\log|x+\sqrt{x^2+4}| - \frac{2}{2}\int \frac{dt}{\sqrt{t}} \)
\( = 3\log|x+\sqrt{x^2+4}| - 2\sqrt{t} + c \)
\( = 3\log|x+\sqrt{x^2+4}| - 2\sqrt{x^2+4} \) ans.

Type: Substitution and then \(\int \frac{1}{\text{Quadratic}} \, dx\)

Question. (a) \( I = \int \frac{\cos x}{\sin^2 x+4\sin x+5} \, dx \)      (b) \( I = \int \frac{\sin(2x)}{\sqrt{\cos^4 x-\sin^2 x+2}} \, dx \)
Answer:
(a) \( I = \int \frac{\cos x}{\sin^2 x+4\sin x+5} \, dx \)
put \( \sin x = t \)
\( \cos x \, dx = dt \)
\( \therefore I = \int \frac{dt}{t^2+4t+5} \)
\( = \int \frac{1}{(t+2)^2-4+5} \, dt \)
\( = \int \frac{1}{(t+2)^2+1} \, dt \)
\( = \tan^{-1}(t+2) + c \)
\( = I = \tan^{-1}(\sin x + 2) + c \) ans.

(b) \( I = \int \frac{\sin(2x)}{\sqrt{\cos^4 x-\sin^2 x+2}} \, dx \)
\( = \int \frac{\sin(2x)}{\sqrt{\cos^4 x-(1-\cos^2 x)+2}} \, dx \)
put \( \cos^2 x = t \)
\( -2\cos x \sin x \, dx = dt \)
\( \sin(2x) \, dx = -dt \)
\( \therefore I = \int \frac{dt}{\sqrt{t^2-(1-t)+2}} \)
\( = -\int \frac{1}{\sqrt{t^2+t+1}} \, dt \)
Proceed Yourself
\( -\log\left| \left(\cos^2 x + \frac{1}{2}\right) + \sqrt{\cos^4 x + \cos^2 x + 1} \right| + c \) ans.

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

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

(b) \( I = \int \frac{1}{x^{2/3}\sqrt{x^{2/3}-4}} \, dx \)
\( I = \int \frac{1}{x^{2/3}\sqrt{(x^{1/3})^2-(2)^2}} \, dx \)
put \( x^{1/3} = t \)
\( \frac{1}{3}x^{-2/3} \, dx = dt \)
\( \frac{1}{x^{2/3}} \, dx = 3 \, dt \)
\( \therefore I = 3\int \frac{dt}{\sqrt{t^2-2^2}} \)
\( = 3\log|t+\sqrt{t^2-4}| + c \)
\( = 3\log|x^{1/3}+\sqrt{x^{2/3}-4}| + c \) ans.

Question. (a) \( I = \int \frac{1}{e^x+1} \, dx \)      (b) \( I = \int \frac{1}{\sqrt{1-e^{2x}}} \, dx \)
Answer:
(a) \( I = \int \frac{1}{e^x+1} \, dx \)
\( = \int \frac{1}{\frac{1}{e^{-x}}+1} \, dx \)
L.C.M
\( = \int \frac{e^{-x}}{1+e^{-x}} \, dx \)
put \( 1+e^{-x} = t \)
\( -e^{-x} \, dx = dt \)
\( e^{-x} \, dx = -dt \)
\( I = -\int \frac{dt}{t} \)
\( = -\log|1+e^{-x}| + c \)
\( = -\log\left|\frac{e^x+1}{e^x}\right| + c \) ans.

(b) \( I = \int \frac{1}{\sqrt{1-e^{2x}}} \, dx \)
\( = \int \frac{1}{\sqrt{1-\frac{1}{e^{-2x}}}} \, dx \)
\( = \int \frac{e^{-x}}{\sqrt{e^{-2x}-1}} \, dx \)
put \( e^{-x} = t \)
\( e^{-x} \, dx = -dt \)
\( \therefore I = -\int \frac{dt}{\sqrt{t^2-1}} \)
\( = -\log|t+\sqrt{t^2-1}| + c \)
\( = -\log|e^{-x}+\sqrt{e^{-2x}-1}| + c \) ans.

Question. (a) \( I = \int \frac{\sin(2x)\cos(2x)}{\sqrt{9-\cos^4(2x)}} \, dx \)      (b) \( I = \int \frac{\sin x+\cos x}{\sqrt{sin(2x)}} \, dx \)      (c) \( I = \int \frac{\sin x-\cos x}{\sqrt{\sin(2x)}} \, dx \)
Answer:
(a) \( I = \int \frac{\sin(2x)\cos(2x)}{\sqrt{9-\cos^4(2x)}} \, dx \)
\( = \int \frac{\sin(2x)\cos(2x)}{\sqrt{9-(\cos^2(2x))^2}} \, dx \)
put \( \cos^2(2x) = t \)
\( -2\cos(2x)\cdot\sin(2x)\cdot2 \, dx = dt \)
\( \cos(2x)\cdot\sin(2x) = \frac{-dt}{4} \)
\( \therefore I = -\frac{1}{4}\int \frac{dt}{\sqrt{3^2-t^2}} \)
\( = -\frac{1}{4}\sin^{-1}\left(\frac{t}{3}\right) + c \)
\( I = -\frac{1}{4}\sin^{-1}\left(\frac{\cos^2(2x)}{3}\right) + c \) ans.

(b) \( I = \int \frac{\sin x+\cos x}{\sqrt{\sin(2x)}} \, dx \)
\( = \int \frac{\sin x+\cos x}{\sqrt{1-1+\sin(2x)}} \, dx \)
\( = \int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2x)}} \, dx \)
\( = \int \frac{\sin x+\cos x}{\sqrt{1-[\sin^2 x+\cos^2 x-2\sin x\cos x]}} \, dx \)
\( = \int \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} \, dx \)
put \( \sin x - \cos x = t \)
\( (\cos x + \sin x) \, dx = dt \)
\( \therefore I = \int \frac{dt}{\sqrt{1-t^2}} \)
\( = \sin^{-1}t + c \)
\( = \sin^{-1}(\sin x - \cos x) + c \) ans.

Type: \(\int \frac{\text{Linear}}{\text{Quadratic}} \, dx\) and \(\int \frac{\text{Linear}}{\sqrt{\text{Quadratic}}} \, dx\) —
Make Derivative of Quadratic in Numerator by Adjustment

Question. (a) \( I = \int \frac{4x+1}{x^2+3x+2} \, dx \)      (b) \( I = \int \frac{3x-2}{1-x-x^2} \, dx \)
Answer:
(a) \( I = \int \frac{4x+1}{x^2+3x+2} \, dx \)
To make \( (2x+3) \)
\( = 2\int \frac{2x+\frac{1}{2}}{x^2+3x+2} \, dx \)
\( = 2\int \frac{2x+\frac{1}{2}+3-3}{x^2+3x+2} \, dx \)
\( = 2\int \frac{(2x+3)-\frac{5}{2}}{x^2+3x+2} \, dx \)
Separate
\( I = 2\int \frac{2x+3}{x^2+3x+2} \, dx - 2 \times \frac{5}{2}\int \frac{1}{x^2+3x+2} \, dx \)
put \( I = 2\int \frac{2x+3}{x^2+3x+2} \, dx - 2 \times \frac{5}{2}\int \frac{1}{x^2+3x+2} \, dx \) in (I)
\( (2x+3) \, dx = dt \)
\( \therefore I = 2\int \frac{dt}{t} - 5\int \frac{1}{\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+2} \, dx \)
\( = 2\log|t| - 5\int \frac{1}{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} \, dx \)
\( = 2\log|x^2+3x+2| - 5 \times \frac{1}{2 \times \frac{1}{2}}\log\left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right| + c \)
\( = 2\log|x^2+3x+2| - 5\log\left|\frac{2x+2}{2x+4}\right| + c \) ans.

(b) \( I = \int \frac{3x-2}{1-x-x^2} \, dx \)
To make \( (-2x-1) \)
\( = 3\int \frac{x-\frac{2}{3}}{1-x-x^2} \, dx \)
\( = -\frac{3}{2}\int \frac{-2x+\frac{4}{3}}{1-x-x^2} \, dx \)
\( = -\frac{3}{2}\int \frac{-2x+\frac{4}{3}-1+1}{1-x-x^2} \, dx \)
\( = -\frac{3}{2}\int \frac{(-2x-1)+\frac{7}{3}}{1-x-x^2} \, dx \)
\( = -\frac{3}{2}\int \frac{(-2x-1)}{1-x-x^2} \, dx - \frac{3}{2} \times \frac{7}{3}\int \frac{1}{1-x-x^2} \, dx \)
put \( 1-x-x^2 = t \)
\( (-2x-1) \, dx = dt \)
\( \therefore I = -\frac{3}{2}\int \frac{dt}{t} - 7\int \frac{1}{1-x-x^2} \, dx \)
\( = -\frac{3}{2}\log|t| - 7\int \frac{1}{-(x^2+x-1)} \, dx \)
\( = -\frac{3}{2}\log|1-x-x^2| - 7\int \frac{1}{-\left[\left(x+\frac{1}{2}\right)^2-\frac{1}{4}-1\right]} \, dx \)
\( = -\frac{3}{2}\log|1-x-x^2| - 7\int \frac{1}{-\left[\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2\right]} \, dx \)
\( = -\frac{3}{2}\log|1-x-x^2| - 7\int \frac{1}{\left(\frac{\sqrt{5}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2} \, dx \)
\( = -\frac{3}{2}\log|1-x-x^2| - 7 \times \frac{1}{2 \times \frac{\sqrt{5}}{2}}\log\left|\frac{\frac{\sqrt{5}}{2}+x+\frac{1}{2}}{\frac{\sqrt{5}}{2}-x-\frac{1}{2}}\right| + c \)
\( = -\frac{3}{2}\log|1-x-x^2| - \frac{7}{\sqrt{5}}\log\left|\frac{\sqrt{5}+1+2x}{\sqrt{5}-1-2x}\right| + c \) ans.

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