CBSE Class 12 Mathematics Integration Worksheet Set 07

CBSE Class 12 Mathematics Integration Worksheet (5). 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. \( I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx \)
Answer:
\( I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx \)
put \( x = t^2 \)
\( dx = 2t \, dt \)
\( \therefore = 2\int \sqrt{\frac{1-t}{1+t}} \cdot t \, dt \)
rationalize
\( = 2\int \sqrt{\frac{1-t}{1+t} \times \frac{1-t}{1-t}} \cdot t \, dt \)
\( = 2\int \frac{1-t}{\sqrt{1-t^2}} \cdot t \, dt \)
\( = 2\int \frac{t-t^2}{\sqrt{1-t^2}} \, dt \)
\( = 2\int \frac{t}{\sqrt{1-t^2}} \, dt - 2\int t^2\sqrt{1-t^2} \, dt \)
put \( 1-t^2 = z \) in (I)
\( -2t \, dt = dz \)
\( t \, dt = -\frac{dz}{2} \)
\( \therefore I = -\frac{2}{2}\int \frac{dz}{\sqrt{z}} + 2\int \frac{-t^2}{\sqrt{1-t^2}} \, dt \)
\( = -2\sqrt{z} + 2\int \frac{1-t^2-1}{\sqrt{1-t^2}} \, dt \)
\( = -2\sqrt{1-t^2} + 2\int \sqrt{1-t^2} - \frac{1}{\sqrt{1-t^2}} \, dt \)
\( = -2\sqrt{1-t^2} + 2\left[ \frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1}(t) - \sin^{-1}(t) \right] + c \)
\( = -2\sqrt{1-t^2} + 2\left[ \frac{t}{2}\sqrt{1-t^2} - \frac{1}{2}\sin^{-1}t \right] + c \)
\( = -2\sqrt{1-t^2} + t\sqrt{1-t^2} - \sin^{-1}t + c \)
replacing t by \( \sqrt{x} \)
\( = I = -2\sqrt{1-x} + \sqrt{x}\sqrt{1-x} - \sin^{-1}\sqrt{x} + c \) ans.

Partial Fraction (Total : Types)

Type : 1 all are linear factors (ax + b)

Question. (a) \( I = \int \frac{2x-1}{(x-1)(x+2)(x-3)} \, dx \)      (b) \( I = \int \frac{x^3}{(x-1)(x-2)} \, dx \)      (c) \( I = \int \frac{x^2}{(x-1)(x-2)(x-3)} \, dx \)      (d) \( I = \int \frac{3+4x-x^2}{(x+2)(x-1)} \, dx \)
Answer:
(a) \( I = \int \frac{2x-1}{(x-1)(x+2)(x-3)} \, dx \)
let \( \frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \)
\( \Rightarrow 2x - 1 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) \)
\( \Rightarrow 2x - 1 = A(x^2 - x - 6) + B(x^2 - 4x + 3) + C(x^2 + x - 2) \)
Comp. the coefficients of \( x^2 \), \( x \) and constant term
\( 0 = A + B + C \Rightarrow C = -A - B \)
\( 2 = -A - 4B + C \Rightarrow 2 = -2A - 5B \)
\( -1 = -6A + 3B - 2C \Rightarrow -1 = -4A - B \)
solving these two equations we get
\( A = \frac{-1}{6}, B = \frac{-1}{3} \) and \( C = \frac{1}{2} \)
\( \therefore I = \int \frac{-1}{6(x-1)} - \frac{1}{3(x+2)} + \frac{1}{2(x-3)} \, dx \)
\( = -\frac{1}{6}\log|x-1| - \frac{1}{3}\log|x+2| + \frac{1}{2}\log|x-3| + c \) ans.

(b) \( I = \int \frac{x^3}{(x-1)(x-2)} \, dx \)
Since degree of Nr > degree of Dr we have to divide
\( \therefore I = \int (x+3) + \frac{7x-6}{(x-1)(x-2)} \, dx \)
\( = \frac{x^2}{2} + 3x + \int \frac{7x-6}{(x-1)(x-2)} \, dx \)
let \( \frac{7x-6}{(x-1)(x-2)} = \frac{A}{x-1} + Bx-2 \) (Correction: \( \frac{B}{x-2} \))
\( \Rightarrow 7x-6 = A(x-2) + B(x-1) \)
Comp. the coefficient of \( x \) and constant term
\( 7 = a + b \)
\( -6 = -2A - B \)
\( 1 = -A \)
\( \therefore B = 8 \)
\( A = -1 \) and \( B = 8 \)
\( \therefore I = \frac{x^2}{2} + 3x + \int \frac{-1}{x-1} + \frac{8}{x-2} \, dx \)
\( I = \frac{x^2}{2} + 3x - \log|x-1| + 8\log|x-2| + c \) ans.

Question. \( I = \int \frac{x}{(x^2+1)(x^2+2)} \, dx \)
Answer:
\( I = \int \frac{x}{(x^2+1)(x^2+2)} \, dx \)
put \( x^2 = t \)
\( x \, dx = \frac{dt}{2} \)
\( \therefore I = \frac{1}{2}\int \frac{dt}{(t+1)(t+2)} \)
let \( \frac{1}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \)
Proceed Yourself
\( \frac{1}{2}[\log|x^2+1| - \log|x^2+2|] + c \) ans.

Question. \( I = \int \frac{1}{\sin x - \sin(2x)} \, dx \)
Answer:
\( I = \int \frac{1}{\sin x - \sin(2x)} \, dx \)
\( = \int \frac{1}{\sin x - 2\sin x \cos x} \, dx \)
\( = \int \frac{1}{\sin x(1-2\cos x)} \, dx \)
multiply and divide by \( \sin x \)
\( = \int \frac{\sin x}{\sin^2 x(1-2\cos x)} \, dx \)
\( = \int \frac{\sin x}{(1-\sin^2 x)(1-\cos x)} \, dx \)
\( = \int \frac{\sin x}{(1-\cos^2 x)(1-2\cos x)} \, dx \)
\( = \int \frac{\sin x}{(1-\cos x)(1+\cos x)(1-2\cos x)} \, dx \)
put \( \cos x = t \)
\( \therefore \sin x \, dx = -dt \)
\( \therefore I = -\int \frac{dt}{(1-t)(1+t)(1-2t)} \)
let \( \frac{1}{(1-t)(1+t)(1-2t)} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{1-2t} \)
\( \Rightarrow 1 = A(1+t)(1-2t) + B(1-t)(1-2t) + C(1-t)(1+t) \)
\( \Rightarrow 1 = A(-2t^2-t+1) + B(2t^2-3t+1) + C(1-t^2) \)
Comp. the coefficient of \( t^2 \), \( t \) and constant term
\( 0 = -2A + 2B - C \Rightarrow C = -2A + 2B \)
\( 0 = -A - 3B \)
\( 1 = A + B + C \)
\( \therefore 1 = -A + 3B \)
\( 0 = -A - 3B \)
\( 1 = -2A \)
\( \therefore A = -\frac{1}{2}, B = \frac{1}{6} \) and \( C = \frac{4}{3} \)
\( \therefore I = -\int \frac{-1}{2(1-t)} + \frac{1}{6(1+t)} + \frac{4}{3(1-2t)} \, dt \)
\( = \left[ +\frac{1}{2}\log|1-t| + \frac{1}{6}\log|1+t| + \frac{4}{3}\log|1-2t|.\left(\frac{-1}{2}\right) \right] + c \)
\( = -\frac{1}{2}\log|1-t| - \frac{1}{6}\log|1+t| + \frac{2}{3}\log|1-2t| + c \)
replacing \( t \)
\( = I = -\frac{1}{2}|1-\cos x| - \frac{1}{6}\log|1+\cos x| + \frac{2}{3}\log|1-\cos x| + c \) ans.

Type : 2 Linear and Quadratic Fraction

Question. (a) \( I = \int \frac{x}{(x-1)(x^2+4)} \, dx \)      (b) \( I = \int \frac{1}{1+x+x^2+x^3} \, dx \)
Answer:
(a) \( I = \int \frac{x}{(x-1)(x^2+4)} \, dx \)
let \( \frac{x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4} \, dx \)
\( \Rightarrow x = A(x^2+4) + (Bx+c)(x-1) \)
\( \Rightarrow x = A(x^2+4) + (Bx^2-Bx+Cx-C) \)
Comp. the coefficient of \( x^2 \), \( x \) and constant term
\( 0 = A + B \)
\( 1 = -B + C \)
\( 0 = 4A - C \)
Solving these equations,
we get \( A = \frac{1}{5}, B = -\frac{1}{5} \) and \( C = \frac{4}{5} \)
\( \therefore I = \int \frac{1}{5(x-1)} + \frac{-\frac{1}{5}x+\frac{4}{5}}{x^2+4} \, dx \)
\( = \frac{1}{5}\int \frac{1}{x-1} \, dx - \frac{1}{5}\int \frac{x}{x^2+4} \, dx + \frac{4}{5}\int \frac{1}{x^2+4} \, dx \)
put \( x^2 + 4 = t \)
\( \therefore x \, dx = \frac{dt}{2} \)
\( \therefore I = \frac{1}{5}\log|x-1| - \frac{1}{10}\int \frac{dt}{t} + \frac{4}{5} \times \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \)
\( I = \frac{1}{5}\log|x-1| - \frac{1}{10}\log|x^2+4| + \frac{2}{3}\tan^{-1}\left(\frac{x}{2}\right) + c \) ans.

(b) \( I = \int \frac{1}{1+x+x^2+x^3} \, dx \)
\( = \int \frac{1}{(1+x)+x^2(1+x)} \, dx \)
\( = \int \frac{1}{(1+x)(1+x^2)} \, dx \)
let \( \frac{1}{(1+x)(1+x^2)} = \frac{A}{1+x} + \frac{Bx+c}{x^2+1} \)
\( 1 = A(x^2+1) + (Bx+C)(x+1) \)
\( 1 = A(x^2+1) + (Bx^2+Bx+Cx+C) \)
Comp. the coefficient of \( x^2 \), \( x \) and constant
\( 0 = A + B \Rightarrow B = -A \)
\( 0 = B + C \Rightarrow 0 = -A + C \)
\( 1 = A + C \qquad 1 = A + C \)
\( 1 = 2C \)
\( C = \frac{1}{2}, A = \frac{1}{2} \) and \( B = -\frac{1}{2} \)
\( \dots I = \frac{1}{2(x+1)} + \frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1} \, dx \)
\( = \frac{1}{2}\int \frac{1}{x+1} \, dx - \frac{1}{2}\int \frac{x}{x^2+1} \, dx + \frac{1}{2}\int \frac{1}{x^2+1} \, dx \)
put \( x^2+1 = t \)
\( dx = \frac{dt}{2} \)
\( = \frac{1}{2}\log|x+1| - \frac{1}{4}\int \frac{dt}{t} + \frac{1}{2}\tan^{-1}x \)
\( I = \frac{-1}{2}\log|x+1| - \frac{1}{4}\log|x^2+1| + \frac{1}{2}\tan^{-1}x + c \) ans.

Question. \( I = \int \frac{x}{x^3-1} \, dx \)
Answer:
\( I = \int \frac{x}{x^3-1} \, dx \)
\( = \int \frac{x}{(1-x)(x^2+x+1)} \, dx \)
let \( \frac{x}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+c}{x^2+x+1} \)
\( x = A(x^2+x+1) + (Bx+c)(x-1) \)
\( x = A(x^2+x+1) + (Bx-Bx^2+cx+c) \)
Comp. the coefficient of \( x^2 \), \( x \) and constant term
\( 0 = A + B \Rightarrow B = -A \)
\( 1 = A - B + C \Rightarrow 1 = 2A + C \)
\( 0 = A - C \Rightarrow 0 = A - C \)
\( 1 = 3A \)
\( A = \frac{1}{3}, B = -\frac{1}{3} \) and \( C = \frac{1}{3} \)
\( \dots I = \int \frac{1}{3(x-1)} + \frac{-\frac{1}{3}x+\frac{1}{3}}{x^2+x+1} \, dx \)
\( = \frac{1}{3}\int \frac{1}{x-1} \, dx - \frac{1}{3}\int \frac{x-1}{x^2+x+1} \, dx \quad \left\{ \text{type } \int \frac{\text{linear}}{\text{quadratic}} \right\} \)
\( = I = \frac{1}{3}\log|x-1| - \frac{1}{3}I \quad \dots (1) \)
where \( I = \int \frac{x-1}{x^2+x+1} \, dx \)
Proceed Yourself and get \( I \)
\( I = \frac{1}{3}\log|x-1| - \frac{1}{6}\log|x^2+x+1| + \frac{1}{\sqrt{3}}\tan^{-1}\left( \frac{2x}{\sqrt{3}} \right) \)

Question. \( I = \int \frac{\tan\theta+\tan^3\theta}{1+\tan^3\theta} \, d\theta \)
Answer:
\( I = \int \frac{\tan\theta+\tan^3\theta}{1+\tan^3\theta} \, d\theta \)
\( = \int \frac{\tan\theta(1+\tan^2\theta)}{1+\tan^3\theta} \, d\theta \)
\( = \int \frac{\tan\theta \cdot \sec^2\theta}{1+\tan^3\theta} \, d\theta \)
put \( \tan\theta = t \)
\( \sec^2\theta \, d\theta = dt \)
\( \dots I = \int \frac{t}{1+t^3} \, dt \)
\( = \int \frac{t}{(1+t)(t^2-t+1)} \, dt \)
Proceed as above Qns. :-
\( -\frac{1}{3}\log|1+\tan\theta| + \frac{1}{6}\log|\tan^2\theta-\tan\theta+1| + \frac{1}{\sqrt{3}}\tan^{-1}\left( \frac{2\tan\theta-1}{\sqrt{3}} \right) \) ans.

Type : 3 Linear and Linear repeating factors

Question. (a) \( I = \int \frac{3x+1}{(x+2)(x-2)^2} \, dx \)      (b) \( I = \int \frac{x^2+x+1}{(x-1)^3} \, dx \)
Answer:
(a) \( I = \int \frac{3x+1}{(x+2)(x-2)^2} \, dx \)
let \( \frac{3x+1}{(x+2)(x-2)^2} = \frac{A}{x+2} + \frac{B}{x-2} + \frac{C}{(x-2)^2} \)
\( 3x + 1 = A(x-2)^2 + B(x-2)(x+2) + C(x+2) \)
\( 3x + 1 = A(x^2 - 4x + 4) + B(x^2 - 4) + C(x+2) \)
Comp. the coefficient of \( x^2 \), \( x \) and constant
\( 0 = A + B \Rightarrow B = -A \)
\( 3 = -4A + C \Rightarrow 3 = -4A + C \)
\( 1 = 4A - 4B + 2C \Rightarrow 1 = 8A + 2C \)
solving these equations, we get
\( A = \frac{-5}{16}, B = \frac{5}{16} \) and \( C = \frac{7}{4} \)
\( \dots I = \frac{-5}{16(x+2)} + \frac{5}{16(x-2)} + \frac{7}{4(x-2)^2} \, dx \)
\( I = \frac{-5}{16}\log|x+2| + \frac{5}{16}\log|x-2| - \frac{7}{4(x-2)} + c \) ans. \( \dots\dots \left\{ \text{Since } \int \frac{1}{x^2} \, dx = \frac{-1}{x} \right\} \)

(b) \( I = \int \frac{x^2+x+1}{(x-1)^3} \, dx \)
let \( \frac{x^2+x+1}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} \)
\( x^2 + x + 1 = A(x-1)^2 + B(x-1) + C \)
\( x^2 + x + 1 = A(x^2 - 2x + 1) + B(x-1) + C \)
Comp. the coefficient of \( x^2 \), \( x \) and constant
\( 1 = A \)
\( 1 = -2A + B \)
\( 1 = A - B + C \)
solving these equation we get \( A = 1, B = 3, C = 3 \)
\( \dots I = \int \frac{1}{x-1} + \frac{3}{(x-1)^2} + \frac{3}{(x-1)^3} \, dx \)
\( = \log|x-1| - \frac{3}{(x-1)} + 3\int (x-1)^{-3} \, dx \)
\( = \log|x-1| - \frac{3}{x-1} + 3 \frac{(x-1)^{-2}}{-2} + c \)
\( \dots I = \log|x-1| - \frac{3}{x-1} - \frac{3}{2(x-1)^2} + c \) ans.

Question. \( I = \int \frac{3x+5}{x^3-x^2-x+1} \, dx \)
Answer:
\( I = \int \frac{3x+5}{x^3-x^2-x+1} \, dx \)
\( = \int \frac{3x+5}{x^2(x-1)-1(x-1)} \, dx \)
\( = \int \frac{3x+5}{(x-1)(x^2-1)} \, dx \)
\( = \int \frac{3x+5}{(x-1)(x+1)(x-1)} \, dx \)
\( = \int \frac{3x+5}{(x+1)(x-1)^2} \, dx \)
let \( \frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \)
\( 3x + 5 = A(x-1)^2 + B(x+1)(x-1) + C(x+1) \)
\( 3x + 5 = A(x^2 - 2x + 1) + B(x^2 - 1) + C(x+1) \)
\( 0 = A + B \)
\( 3 = -2A + C \)
\( 5 = A - B + C \)
solving these equation, we get \( A = \frac{1}{2}, B = \frac{-1}{2}, C = 4 \)
\( \dots I = \int \frac{1}{2(x+1)} - \frac{1}{2(x-1)} + \frac{4}{(x-1)^2} \, dx \)
\( I = \frac{1}{2}\log|x+1| - \frac{1}{2}\log|x-1| - \frac{4}{x-1} + c \) ans.

Type : 4 Even Power of x let \(x^2\) = y (temp.)

Question. (a) \( I = \int \frac{x^2}{(x^2+1)(x^2+4)} \, dx \)      (b) \( I = \int \frac{1}{(x^4-1)} \, dx \)
Answer:
(a) \( I = \int \frac{x^2}{(x^2+1)(x^2+4)} \, dx \)
let \( x^2 = y \)
\( \therefore \frac{x^2}{(x^2+1)(x^2+4)} = \frac{y}{(y+1)(y+4)} \)
let \( \frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4} \)
\( y = A(y+4) + B(y+1) \)
Comp. coefficient of \( y \) and constant
\( 1 = A + B \)
\( 0 = 4A + B \)
\( 1 = -3A \)
\( A = \frac{-1}{3}, B = \frac{4}{3} \)
\( \dots I = \int \frac{-1}{3(x^2+1)} + \frac{4}{3(x^2+4)} \, dx \)
\( = \frac{-1}{3}\int \frac{1}{(x^2+1)} \, dx + \frac{4}{3}\int \frac{1}{x^2+(2)^2} \, dx \)
\( = \frac{-1}{3}\tan^{-1}x + \frac{4}{3} \times \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \) ans.

(b) \( I = \int \frac{1}{(x^4-1)} \, dx \)
\( = \int \frac{1}{(x^2+1)(x^2-1)} \, dx \)
let \( x^2 = y \)
\( \dots \frac{1}{(x^2+1)(x^2-1)} = \frac{1}{(y+1)(y-1)} \)
let \( \frac{1}{(y+1)(y-1)} = \frac{A}{y+1} + \frac{B}{y-1} \)
\( 1 = A(y-1) + B(y+1) \)
Comp.
\( 0 = A + B \)
\( 1 = -A + B \)
\( 1 = 2B \)
\( \dots B = \frac{1}{2} \) and \( A = \frac{-1}{2} \)
\( \dots I = \int \frac{-1}{2(x^2+1)} + \frac{1}{2(x^2-1)} \, dx \)
\( = \frac{-1}{2}\int \frac{1}{x^2+1} \, dx + \frac{1}{2}\int \frac{1}{x^2-1} \, dx \)
\( = \frac{-1}{2}\tan^{-1}x + \frac{1}{2} \times \frac{1}{2 \times 1}\log\left|\frac{x-1}{x+1}\right| + c \)
\( = \frac{-1}{2}\tan^{-1}x + \frac{1}{4}\log\left|\frac{x-1}{x+1}\right| + c \) ans.

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