CBSE Class 12 Mathematics Integration Worksheet Set 09

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Integration (Indefinite Integrals)

Question. (a) \( I = \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} \, dx \)      (b) \( I = \int \frac{\sin x}{\sin(4x)} \, dx \)
Answer:
(a) \( I = \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} \, dx \)
since degree of N = degree, D = Divide
\( \therefore I = \int \left( 1 + \frac{-4x^2-10}{(x^2+3)(x^2+4)} \right) \, dx \)
\( I = x - \int \frac{4x^2 + 10}{(x^2 + 3)(x^2 + 4)} \, dx \)
let \( x^2 = y \)
\( \therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{4y+10}{(y+3)(y+4)} \)
let \( \frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4} \)
\( 4y + 10 = A(y + 4) + B(y + 3) \)
Comp. the coefficient of \( y \) and constant
\( 4 = A + B \)
\( 10 = 4A + 3B \)
solving these equations, we get \( A = -2, B = 6 \)
\( \therefore I = x - \int \left( \frac{-2}{x^2+3} + \frac{6}{x^2+4} \right) \, dx \)
\( I = x + 2\int \frac{1}{x^2 + (\sqrt{3})^2} \, dx - 6\int \frac{1}{x^2 + 2^2} \, dx \)
\( I = x + 2 \times \frac{1}{\sqrt{3}}\tan^{-1} \left( \frac{x}{\sqrt{3}} \right) - \frac{6}{2}\tan^{-1} \left( \frac{x}{2} \right) + c \)      ans.

(b) \( I = \int \frac{\sin x}{\sin(4x)} \, dx \)
\( = \int \frac{\sin x}{2\sin(2x)\cos(2x)} \, dx \)
\( = \int \frac{\sin x}{4\sin x \cos x \cos(2x)} \, dx \)
\( = \frac{1}{4}\int \frac{1}{\cos x(1-2\sin^2 x)} \, dx \qquad \dots\dots \{\cos(2\theta) = 1 - 2\sin^2\theta\} \)
multiply and divide by \( \cos x \)
\( = \frac{1}{4}\int \frac{\cos x}{\cos^2 x(1-2\sin^2 x)} \, dx \)
\( = \frac{1}{4}\int \frac{\cos x}{(1-\sin^2 x)(1-2\sin^2 x)} \, dx \)
put \( \sin x = t \)
\( \therefore \cos x \, dx = dt \)
\( \therefore I = \frac{1}{4}\int \frac{1}{(1-t^2)(1-2t^2)} \, dt \)
let \( t^2 = y \)
\( \therefore \frac{1}{(1-t^2)(1-2t^2)} = \frac{1}{(1-y)(1-2y)} \)
let \( \frac{1}{(1-y)(1-2y)} = \frac{A}{1-y} + \frac{B}{1-2y} \)
\( 1 = A(1 - 2y) + B(1 - y) \)
\( 0 = -2A - B \)
\( 1 = A + B \)
\( 1 = -A \)
\( \therefore A = -1 \) and \( B = 2 \)
\( \therefore I = \frac{1}{4}\int \left( \frac{-1}{1-y} + \frac{2}{1-2y} \right) \, dt \)
\( = \frac{1}{4}\int \left( \frac{-1}{1-t^2} + \frac{2}{1-2t^2} \right) \, dt \)
\( = \frac{-1}{4}\int \frac{1}{1-t^2} \, dt + \frac{1}{2}\int \frac{1}{1-2t^2} \, dt \)
\( = \frac{-1}{4}\int \frac{1}{1-t^2} \, dt + \frac{1}{4}\int \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^2 - t^2} \, dt \)
\( = \frac{-1}{4} \times \frac{1}{2 \times 1}\log\left| \frac{1+t}{1-t} \right| + \frac{1}{4} \times \frac{1}{2 \times \frac{1}{\sqrt{2}}}\log\left| \frac{\frac{1}{\sqrt{2}}+t}{\frac{1}{\sqrt{2}}-t} \right| + c \)
\( I = \frac{-1}{8}\log\left| \frac{1+\sin x}{1-\sin x} \right| + \frac{1}{4\sqrt{2}}\log\left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right| + c \)      ans.

\( x^4 \) type

Form: \( \int \frac{x^2+1}{x^4+1} \, dx \) or \( \int \frac{x^2-1}{x^4+1} \, dx \)

• \( D^r \to x^4 \)
• No odd power of \( x \)
• \( D^r \to \text{constant term (+ve)} \)
• \( D^r \text{ constant} = (N^r \text{ constant})^2 \)

Procedure:

• Divide \( N^r \) & \( D^r \) by \( x^2 \)
• In \( N^r \) we get either \( 1 + \frac{1}{x^2} \, dx \) or \( 1 - \frac{1}{x^2} \, dx \)
• In \( D^r \) put \( x - \frac{1}{x} = t \) (or) \( x + \frac{1}{x} = t \)
• Use identities:
  • \( a^2 + b^2 = (a + b)^2 - 2ab \)
  • \( a^2 + b^2 = (a - b)^2 + 2ab \)

Question. (a) \( I = \int \frac{x^2+1}{x^4+1} \, dx \)      (b) \( I = \int \frac{x^2-4}{x^4-16} \, dx \)
Answer:
(a) \( I = \int \frac{x^2+1}{x^4+1} \, dx \)
Divide \( N^r \) & \( D^r \) by \( x^2 \)
\( \therefore I = \int \frac{1+1/x^2}{x^2+\frac{1}{x^2}} \, dx \)
\( = \int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2 + 2} \, dx \)
put \( x - \frac{1}{x} = t \)
\( \left(1 + \frac{1}{x^2}\right) dx = dt \)
\( I = \int \frac{dt}{t^2 + 2} \)
\( = \frac{1}{\sqrt{2}}\tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + c \)
\( = \frac{1}{\sqrt{2}}\tan^{-1} \left( \frac{x-\frac{1}{x}}{\sqrt{2}} \right) + c \)
\( I = \frac{1}{\sqrt{2}}\tan^{-1} \left( \frac{x^2-1}{\sqrt{2}x} \right) + c \)      ans.

(b) \( I = \int \frac{x^2-4}{x^4-16} \, dx \)
Divide \( N^r \) & \( D^r \) by \( x^2 \)
\( = \int \frac{1-\frac{4}{x^2}}{x^2-\frac{16}{x^2}} \, dx \)
\( = \int \frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2 - 8} \, dx \)
put \( x + \frac{4}{x} = t \)
\( \left( 1 - \frac{4}{x^2} \right) dx = dt \)
\( \therefore I = \int \frac{dt}{t^2 - (2\sqrt{2})^2} \)
\( = \frac{1}{2 \times 2\sqrt{2}}\log\left| \frac{t-2\sqrt{2}}{t+2\sqrt{2}} \right| + c \)
\( = \frac{1}{4\sqrt{2}}\log\left| \frac{x+\frac{4}{x}-2\sqrt{2}}{x+\frac{4}{x}+2\sqrt{2}} \right| + c \)
\( = \frac{1}{4\sqrt{2}}\log\left| \frac{x^2-2\sqrt{2}x+4}{x^2+2\sqrt{2}x+4} \right| + c \)      ans.

Question. (a) \( I = \int \frac{x^2}{x^4+x^2+1} \, dx \)      (b) \( I = \int \frac{1}{x^4+1} \, dx \)
Answer:
(a) \( I = \int \frac{x^2}{x^4+x^2+1} \, dx \)      (double set)
Divide \( N^r \) & \( D^r \) by \( x^2 \)
\( I = \int \frac{1}{x^2 + 1 + \frac{1}{x^2}} \, dx \)
\( = \frac{1}{2}\int \frac{2}{x^2+\frac{1}{x^2}+1} \, dx \)      (adjustment)
\( = \frac{1}{2}\int \frac{1+1+\frac{1}{x^2}-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} \, dx \)      (adjustment)
\( = \frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} \, dx + \frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} \, dx \)
\( = \frac{1}{2}\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2 + 2+1} \, dx + \frac{1}{2}\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2 - 2+1} \, dx \)
put \( x - \frac{1}{x} = t \)      put \( x + \frac{1}{x} = z \)
\( \therefore \left(1 + \frac{1}{x^2}\right) dx = dt \qquad \therefore \left(1 - \frac{1}{x^2}\right) dx = dz \)
\( \therefore I = \frac{1}{2}\int \frac{dt}{t^2+3} + \frac{1}{2}\int \frac{dz}{z^2-1} \)
\( = \frac{1}{2} \times \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + \frac{1}{2} \times \frac{1}{2\times 1}\log\left|\frac{z-1}{z+1}\right| + c \)
\( = \frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right) + \frac{1}{4}\log\left| \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} \right| + c \)
\( = \frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right) + \frac{1}{4}\log\left| \frac{x^2-x+1}{x^2+x+1} \right| + c \)      ans.

(b) \( I = \int \frac{1}{x^4+1} \, dx \)      (double set)
Divide \( N^r \) & \( D^r \) by \( x^2 \)
\( I = \int \frac{\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx \)
\( = \frac{1}{2}\int \frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}} \, dx \)
\( = \frac{1}{2}\int \frac{1+\frac{1}{x^2}-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}} \, dx \)
\( = \frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}} \, dx - \frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}} \, dx \)
\( = \frac{1}{2}\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2 + 2} \, dx - \frac{1}{2}\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2 - 2} \, dx \)
put \( x - \frac{1}{x} = t \)      put \( x + \frac{1}{x} = z \)
\( \left(1 + \frac{1}{x^2}\right) dx = dt \qquad \left(1 - \frac{1}{x^2}\right) dx = dz \)
\( \therefore I = \frac{1}{2}\int \frac{dt}{t^2+2} - \frac{1}{2}\int \frac{dz}{z^2-2} \)
\( = \frac{1}{2} \times \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right) - \frac{1}{2} \times \frac{1}{2\times \sqrt{2}}\log\left| \frac{z-\sqrt{2}}{z+\sqrt{2}} \right| + c \)
\( = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) - \frac{1}{4\sqrt{2}}\log\left| \frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}} \right| + c \)
\( = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) - \frac{1}{4\sqrt{2}}\log\left| \frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1} \right| + c \)      ans.

Question. (a) \( I = \int \left(\sqrt{\tan\theta} + \sqrt{\cot\theta}\right) \, d\theta \)      (b) \( I = \int \sqrt{\cot\theta} \, d\theta \)
Answer:
(a) \( I = \int \left(\sqrt{\tan\theta} + \sqrt{\cot\theta}\right) \, d\theta \)
\( = \int \left(\sqrt{\tan\theta} + \frac{1}{\sqrt{\tan\theta}}\right) \, d\theta \)
\( = \int \frac{\tan\theta+1}{\sqrt{\tan\theta}} \, d\theta \)
put \( \tan\theta = t^2 \)
\( \sec^2\theta \, d\theta = 2t \, dt \)
\( d\theta = \frac{2t \, dt}{\sec^2\theta} \)
\( d\theta = \frac{2t \, dt}{\tan^2\theta+1} \)
\( d\theta = \frac{2t \, dt}{t^4+1} \qquad \dots\dots \{\because \tan\theta = t^2\} \)
\( \therefore I = \int \frac{t^2+1}{t} \cdot \frac{2t \, dt}{t^4+1} \)
\( = I = 2\int \frac{t^2+1}{t^4+1} \, dt \)      (single set)
Proceed yourself
\( I = \sqrt{2}\tan^{-1}\left( \frac{\tan\theta-1}{\sqrt{2\tan\theta}} \right) + c \)      ans.

(b) \( I = \int \sqrt{\cot\theta} \, d\theta \)      (double set)
put \( \cot\theta = t^2 \)
\( \therefore -\csc^2\theta \, d\theta = 2t \, dt \)
\( d\theta = \frac{-2t \, dt}{\csc^2\theta} \)
\( d\theta = \frac{-2t \, dt}{\cot^2\theta+1} \)
\( d\theta = \frac{-2t \, dt}{t^4+1} \qquad \dots\dots \{\because \cot\theta = t^2\} \)
\( \therefore I = \int t \cdot \left( \frac{-2t \, dt}{t^4+1} \right) \)
\( = -2\int \frac{t^2}{t^4+1} \, dt \)
Divide \( N^r \) & \( D^r \) by \( t^2 \)
\( = -2\int \frac{1}{t^2 + \frac{1}{t^2}} \, dt \)
\( = \frac{-2}{2}\int \frac{2}{t^2+\frac{1}{t^2}} \, dt \)
\( = -\int \frac{1+1}{t^2+\frac{1}{t^2}} \, dt \)
\( = -\int \frac{1+1+\frac{1}{t^2}-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt \)
\( = -\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt \)
\( = -\int \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2} \, dt - \int \frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2} \, dt \)
put \( t - \frac{1}{t} = u \) and put \( t + \frac{1}{t} = v \)
\( \left(1 + \frac{1}{t^2}\right) dt = du \qquad \left(1 - \frac{1}{t^2}\right) dt = dv \)
\( \therefore I = -\int \frac{du}{u^2+(\sqrt{2})^2} - \int \frac{dv}{v^2-(\sqrt{2})^2} \)
\( = \frac{-1}{\sqrt{2}}\tan^{-1}\left( \frac{u}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\log\left| \frac{v-\sqrt{2}}{v+\sqrt{2}} \right| + c \)
\( = \frac{-1}{\sqrt{2}}\tan^{-1}\left( \frac{t-\frac{1}{t}}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\log\left| \frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}} \right| + c \)
\( = \frac{-1}{\sqrt{2}}\tan^{-1}\left( \frac{t^2-1}{\sqrt{2}t} \right) - \frac{1}{2\sqrt{2}}\log\left| \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} \right| + c \)
replacing \( t \) by \( \sqrt{\cot\theta} \)
\( I = \frac{-1}{\sqrt{2}}\tan^{-1}\left( \frac{\cot\theta-1}{\sqrt{2\cot\theta}} \right) - \frac{1}{2\sqrt{2}}\log\left| \frac{\cot\theta-\sqrt{2\cot\theta}+1}{\cot\theta+\sqrt{2\cot\theta}+1} \right| + c \)      ans.

Question. \( I = \int \frac{1}{\sin^4 x + \cos^4 x} \, dx \)
Answer:
\( I = \int \frac{1}{\sin^4 x + \cos^4 x} \, dx \)
Divide \( N^r \) & \( D^r \) by \( \cos^4 x \)
\( I = \int \frac{\sec^4 x}{\tan^4 x + 1} \, dx \)
\( = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^4 x+1} \, dx \)
\( = \int \frac{(\tan^2 x+1)\sec^2 x}{\tan^4 x+1} \, dx \)
put \( \tan x = t \)
\( \therefore \sec^2 x \, dx = dt \)
\( I = \int \frac{t^2+1}{t^4+1} \, dt \)      (single set)
Proceed Yourself
\( I = \frac{1}{\sqrt{2}}\tan^{-1}\left( \frac{\tan^2 x - 1}{\sqrt{2}\tan x} \right) + c \)      ans.

Type:

• \( \int \frac{\phi(x)}{\text{linear}\sqrt{\text{linear}}} \, dx \); put Linear = \( t^2 \)
• \( \int \frac{\phi(x)}{\text{quadratic}\sqrt{\text{linear}}} \, dx \); put Linear = \( t^2 \)
• \( \int \frac{\phi(x)}{\text{Linear}\sqrt{\text{quadratic}}} \, dx \); put Linear = \( \frac{1}{t} \)

Question. (a) \( I = \int \frac{1}{(x-3)\sqrt{x+1}} \, dx \)      (b) \( I = \int \frac{1}{(x-1)\sqrt{2x+3}} \, dx \)
Answer:
(a) \( I = \int \frac{1}{(x-3)\sqrt{x+1}} \, dx \)
put \( x + 1 = t^2 \)
\( dx = 2t \, dt \)
\( \dots I = \int \frac{1}{(x-3)t} \cdot 2t \, dt \)
\( = 2\int \frac{1}{(t^2-1-3)} \, dt \qquad \dots\dots \{\because x = t^2 - 1\} \)
\( = 2\int \frac{1}{t^2-4} \, dt \)
\( = 2 \times \frac{1}{2 \times 2}\log\left| \frac{t-2}{t+2} \right| + c \)
\( I = \frac{1}{2}\log\left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right| + c \)

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

Question. (a) \( I = \int \frac{1}{(x^2-4)\sqrt{x+1}} \, dx \)      (b) \( I = \int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}} \, dx \)
Answer:
(a) \( I = \int \frac{1}{(x^2-4)\sqrt{x+1}} \, dx \)
put \( x + 1 = t^2 \)
\( dx = 2t \, dt \)
\( \therefore I = \int \frac{2t}{(x^2-4)t} \, dt \)
\( = 2\int \frac{1}{(t^2-1)^2-4} \, dt \)
\( = 2\int \frac{1}{t^4-2t^2+1-4} \, dt \)
\( = 2\int \frac{1}{t^4-2t^2-3} \, dt \)
\( = 2\int \frac{1}{(t^2-3)(t^2+1)} \, dt \)
Partial fraction: type 4
let \( t^2 = y \)
Proceed yourself
\( I = \frac{1}{4\sqrt{3}}\log\left| \frac{\sqrt{x+1}-\sqrt{3}}{\sqrt{x+1}+\sqrt{3}} \right| - \frac{1}{2}\tan^{-1}(\sqrt{x+1}) + c \)      ans.

(b) \( I = \int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}} \, dx \)
put \( x+1 = t^2 \)
\( dx = 2t \, dt \)
\( \therefore I = \int \frac{x+2}{(x^2+3x+3)} \cdot \frac{2t}{t} \, dt \)
\( = 2\int \frac{(t^2-1)+2}{(t^2-1)^2+3(t^2-1)+3} \, dt \qquad \dots\dots \{\because x = t^2 - 1\} \)
\( = 2\int \frac{t^2+1}{t^4-2t^2+1+3t^2-3+3} \, dt \)
\( = 2\int \frac{t^2-1}{t^4+t^2+1} \, dt \)      (single set)
\( x^4 \) type single set
Proceed Yourself
\( \frac{2}{\sqrt{3}}\tan^{-1}\left( \frac{x}{\sqrt{3(x+1)}} \right) + c \)      ans.

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

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