CBSE Class 12 Mathematics Integration Worksheet Set 08

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Integration (Indefinite Integrals)

Type: \( \int e^{ax} \sin(bx + c) \, dx \), \( \int e^{ax} \cos(bx + c) \, dx \)

I repeats of the two types by Parts

Question. (a) \( I = \int e^{2x} \cdot \cos(3x) \, dx \)      (b) \( I = \int e^{ax} \cdot \sin(dx + c) \, dx \)
Answer:
(a) \( I = \int e^{2x} \cdot \cos(3x) \, dx \)
\( = \cos(3x) \cdot \frac{e^{2x}}{2} - \int (-3\sin(3x)) \cdot \frac{e^{2x}}{2} \, dx \)
\( = \frac{e^{2x}}{2} \cdot \cos(3x) + \frac{3}{2}\int \sin(3x) \cdot e^{2x} \, dx \)
\( = \frac{e^{2x}}{2} \cdot \cos(3x) + \frac{3}{2} \left[ \sin(3x) \cdot \frac{e^{2x}}{2} - \int 3\cos(3x) \cdot \frac{e^{2x}}{2} \, dx \right] \)
\( I = \frac{e^{2x}}{2} \cdot \cos(3x) + \frac{3}{2} \left[ \frac{e^{2x}}{2} \cdot \sin(3x) - \frac{3}{2} I \right] \)
\( I = \frac{e^{2x}}{2} \cdot \cos(3x) + \frac{3}{4} e^{2x} \cdot \sin(3x) - \frac{9}{4} I \)
\( I + \frac{9}{4} I = \frac{e^{2x}}{4} [2\cos(3x) + 3\sin(3x)] \)
\( \frac{13I}{4} = \frac{e^{2x}}{4} [2\cos(3x) + 3\sin(3x)] + c \)
\( \therefore I = \frac{e^{2x}}{13} [2\cos(3x) + 3\sin(3x)] + c \)      ans.

(b) \( I = \int e^{ax} \cdot \sin(bx + c) \, dx \)
\( = \sin(bx + c) \cdot \frac{e^{ax}}{a} - \int b\cos(bx + c) \cdot \frac{e^{ax}}{a} \, dx \)
\( = \frac{e^{ax}}{a} \sin(bx + c) - \frac{b}{a}\int e^{ax} \cdot \cos(bx + c) \, dx \)
\( = \frac{e^{ax}}{a} \cdot \sin(bx + c) - \frac{b}{a} \left[ \cos(bx + c) \cdot \frac{e^{ax}}{a} - \int (-b\sin(bx + c)) \cdot \frac{e^{ax}}{a} \, dx \right] \)
\( = \frac{e^{ax}}{a} \cdot \sin(bx + c) - \frac{b}{a} \left[ \frac{e^{ax}}{a} \cdot \cos(bx + c) + \frac{b}{a}\int e^{ax} \cdot \sin(bx + c) \, dx \right] \)
\( = \frac{e^{ax}}{a} \cdot \sin(bx + c) - \frac{b}{a} \left[ \frac{e^{ax}}{a} \cos(bx + c) + \frac{b}{a} I \right] \)
\( I = \frac{e^{ax}}{a} \cdot \sin(bx + c) - \frac{b^2}{a^2} e^{ax} \cdot \cos(bx + c) - \frac{b^2}{a^2} I \)
\( I + \frac{b^2}{a^2} I = \frac{e^{ax}}{a^2} [a \sin(bx + c) - b \cos(bx + c)] + c \)
\( I \left( \frac{a^2 + b^2}{a^2} \right) = \frac{e^{ax}}{a^2} [a \sin(bx + c) - b\cos(bx + c)] + c \)
\( \therefore I = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx + c) - b \cos(bx + c)] + c \)      ans.

Question. \( I = \int e^x \cdot \cos^2 x \, dx \)
Answer:
\( I = \int e^x \cdot \cos^2 x \, dx \)
\( = \int e^x \cdot \left\{ \frac{1+\cos(2x)}{2} \right\} \, dx \)
\( = \frac{1}{2}\int e^x + e^x \cdot \cos(2x) \, dx \)
\( I = \frac{1}{2}\int e^x \, dx + \frac{1}{2}\int e^x \cdot \cos(2x) \, dx \)
\( I = \frac{1}{2}\int e^x \, dx + \frac{1}{2} I_1 \)
where \( I_1 = \int e^x \cdot \cos(2x) \, dx \)
\( = \cos(2x) \cdot e^x - \int -2\sin(2x) \cdot e^x \, dx \)
\( = e^x \cdot \cos(2x) + 2\int e^x \cdot \sin(2x) \, dx \)
\( = e^x \cdot \cos(2x) + 2 [ e^x \cdot \sin(2x) - 2\int \cos(2x) \cdot e^x \, dx ] \)
\( I_1 = e^x \cos(2x) + 2 e^x \sin(2x) - 4 I_1 \)
\( 5 I_1 = e^x [\cos(2x) + 2\sin(2x)] \)
\( I_1 = \frac{e^x}{5} [\cos(2x) + 2\sin(2x)] + c \)
\( \therefore I = \frac{1}{2} e^x + \frac{1}{2} \left[ \frac{e^x}{5} \cdot (\cos(2x) + 2\sin(2x)) \right] + c \)      ans.

Type: \( I = \int e^x ( f(x) + f'(x) ) \, dx \)

\( I = \int e^x \cdot f(x) \, dx + \int e^x \cdot f'(x) \, dx \)
\( = f(x) \cdot e^x - \int f'(x) \cdot e^x \, dx + \int e^x \cdot f'(x) \, dx \)
\( I = e^x \cdot f(x) + c \)

Question. (a) \( I = \int e^x \left( \frac{2+\sin(2x)}{1+\cos(2x)} \right) \, dx \)      (b) \( I = \int e^x \left( \frac{1-\sin x}{1-\cos x} \right) \, dx \)      (c) \( I = \int e^{2x} \left( \frac{1+\sin(2x)}{1+\cos(2x)} \right) \, dx \)
Answer:
(a) \( I = \int e^x \left( \frac{2+\sin(2x)}{1+\cos(2x)} \right) \, dx \)
\( = \int e^x \left[ \frac{2+2\sin x \cdot \cos x}{2\cos^2 x} \right] \, dx \)
\( = \int e^x \left[ \frac{2}{2\cos^2 x} + \frac{2\sin x \cdot \cos x}{2\cos^2 x} \right] \, dx \)
\( = \int e^x (\sec^2 x + \tan x) \, dx \qquad \dots\dots \{f'(x) = \sec^2 x, f(x) = \tan x\} \)
\( = \int e^x \cdot \tan x \, dx + \int e^x \sec^2 x \, dx \)
\( = \tan x \cdot e^x - \int \sec^2 x \cdot e^x \, dx + \int e^x \sec^2 x \, dx \)
\( = e^x \cdot \tan x + c \)      ans.

(b) \( I = \int e^x \left( \frac{1-\sin x}{1-\cos x} \right) \, dx \)
\( = \int e^x \left[ \frac{1 - 2\sin\frac{x}{2} \cdot \cos\frac{x}{2}}{2\sin^2\frac{x}{2}} \right] \, dx \)
\( = \int e^x \left[ \frac{1}{2\sin^2\frac{x}{2}} - \frac{2\sin\frac{x}{2} \cdot \cos\frac{x}{2}}{2\sin^2\frac{x}{2}} \right] \, dx \)
\( = \int e^x \left[ \frac{1}{2}\csc^2\left(\frac{x}{2}\right) - \cot\left(\frac{x}{2}\right) \right] \, dx \qquad \dots\dots \{f'(x) = \frac{1}{2}\csc^2\left(\frac{x}{2}\right), f(x) = -\cot\left(\frac{x}{2}\right)\} \)
\( = -\int e^x \cdot \cot\left(\frac{x}{2}\right) \, dx + \frac{1}{2}\int e^x \cdot \csc^2\left(\frac{x}{2}\right) \, dx \)
\( = -\left[ \cot\left(\frac{x}{2}\right) \cdot e^x - \int -\frac{1}{2}\csc^2\left(\frac{x}{2}\right) \cdot e^x \, dx \right] + \frac{1}{2}\int e^x \cdot \csc^2\left(\frac{x}{2}\right) \, dx \)
\( = -e^x \cot\left(\frac{x}{2}\right) - \frac{1}{2}\int e^x \cdot \csc^2\left(\frac{x}{2}\right) \, dx + \frac{1}{2}\int e^x \cdot \csc^2\left(\frac{x}{2}\right) \, dx \)
\( = I = -e^x \cdot \cot\left(\frac{x}{2}\right) + c \)      ans.

(c) \( I = \int e^{2x} \left( \frac{1+\sin(2x)}{1+\cos(2x)} \right) \, dx \)
\( = \int e^{2x} \left( \frac{1+2\sin x \cos x}{2\cos^2 x} \right) \, dx \)
\( = \int e^{2x} \cdot \left( \frac{1}{2}\sec^2 x + \tan x \right) \, dx \)
\( = \int e^{2x} \cdot \tan x \, dx + \frac{1}{2}\int e^{2x} \cdot \sec^2 x \, dx \)
\( = \tan x \cdot \frac{e^{2x}}{2} - \int \sec^2 x \cdot \frac{e^{2x}}{2} \, dx + \frac{1}{2}\int e^{2x} \cdot \sec^2 x \, dx \)
\( I = \frac{1}{2} e^{2x} \cdot \tan x + c \)      ans.

Question. (a) \( I = \int e^x - \frac{x}{(x+1)^2} \, dx \)      (b) \( I = \int e^x \left( \frac{x-4}{(x-2)^3} \right) \, dx \)
Answer:
(a) \( I = \int e^x \cdot \frac{x}{(x+1)^2} \, dx \)
\( = \int e^x \left[ \frac{x+1-1}{(x+1)^2} \right] \, dx \)
\( = \int e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] \, dx \qquad \dots\dots \{f(x) = \frac{1}{x+1}, f'(x) = -\frac{1}{(x+1)^2}\} \)
\( = \int e^x \cdot \frac{1}{x+1} \, dx - \int e^x \cdot \frac{1}{(x+1)^2} \, dx \)
\( = \frac{1}{x+1} \cdot e^x + \int \frac{1}{(x+1)^2} \cdot e^x \, dx - \int \frac{1}{(x+1)^2} \cdot e^x \, dx \)
\( I = e^x \cdot \frac{1}{x+1} + c \)      ans.

(b) \( I = \int e^x \left( \frac{x-4}{(x-2)^3} \right) \, dx \)
\( = \int e^x \left( \frac{x-4}{(x-2)^3} \right) \, dx \)
\( = \int e^x \left[ \frac{1}{(x-2)^2} - \frac{2}{(x-2)^3} \right] \, dx \qquad \dots\dots \{f(x) = \frac{1}{(x-2)^2}, f'(x) = -\frac{2}{(x-2)^3}\} \)
Proceed Yourself
\( e^x \cdot \frac{1}{(x-2)^2} + c \)      ans.

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

Question. (a) \( I = \int \frac{\log x}{(\log x+1)^2} \, dx \)      (b) \( I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] \, dx \)
Answer:
(a) \( I = \int \frac{\log x}{(\log x+1)^2} \, dx \)
put \( \log x = t \)
\( x = e^t \)
\( dx = e^t \, dt \)
\( \therefore I = \int \frac{t}{(t+1)^2} \cdot e^t \, dt \)
\( = \int e^t \left[ \frac{t+1-1}{(t+1)^2} \right] \, dt \)
\( = \int e^t \left[ \frac{1}{t+1} - \frac{1}{(t+1)^2} \right] \, dt \)
\( = e^t \cdot \frac{1}{t+1} + c \)
replacing \( t \):
\( = x \cdot \frac{1}{\log x+1} + c \)      ans.

(b) \( I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] \, dx \)
put \( \log x = t \)
\( x = e^t \)
\( dx = e^t \, dt \)
\( \therefore I = \int \left( \log t + \frac{1}{t^2} \right) \cdot e^t \, dt \)
adjustment:
\( = \int e^t \left[ \log t + \frac{1}{t} - \frac{1}{t} + \frac{1}{t^2} \right] \, dt \)
\( = \int e^t \left( \log t + \frac{1}{t} \right) \, dt - \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) \, dt \)
\( = \left[ \int e^t \log t \, dt + \int e^t \cdot \frac{1}{t} \, dt \right] - \left[ \int e^t \cdot \frac{1}{t} \, dt - \int e^t \cdot \frac{1}{t^2} \, dt \right] \)
\( = \left[ \log t \cdot e^t - \int \frac{1}{t} \cdot e^t \, dt + \int e^t \cdot \frac{1}{t} \, dt \right] - \left[ \frac{1}{t} \cdot e^t + \int \frac{1}{t^2} \cdot e^t \, dt - \int e^t \cdot \frac{1}{t^2} \, dt \right] \)
\( = \log t \cdot e^t - \frac{1}{t} \cdot e^t + c \)
\( = e^t \left( \log t - \frac{1}{t} \right) + c \)
\( I = x \left( \log(\log x) - \frac{1}{\log x} \right) + c \)      ans.

Question. (a) \( I = \int e^{-x/2} \frac{\sqrt{1-\sin x}}{1+\cos x} \, dx \)      (b) \( I = \int e^{2x} (-\sin x + 2\cos x) \, dx \)
Answer:
(a) \( I = \int e^{-x/2} \frac{\sqrt{1-\sin x}}{1+\cos x} \, dx \)
\( = \int e^{-x/2} \frac{\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}}{2\cos^2\left(\frac{x}{2}\right)} \, dx \)
\( = \int e^{-x/2} \frac{\sqrt{\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2}}{2\cos^2\frac{x}{2}} \, dx \)
\( = \int e^{-x/2} \frac{\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} \, dx \)
\( = \int e^{-x/2} \left[ \frac{1}{2}\tan\frac{x}{2} \cdot \sec\frac{x}{2} - \frac{1}{2}\sec\left(\frac{x}{2}\right) \right] \, dx \)
\( = -\frac{1}{2}\int e^{-x/2} \sec\left(\frac{x}{2}\right) \, dx + \frac{1}{2}\int e^{-x/2} \sec\frac{x}{2} \cdot \tan\frac{x}{2} \, dx \)
\( = -\frac{1}{2} \left[ \sec\left(\frac{x}{2}\right) \cdot e^{-x/2}(-2) - \int \sec\left(\frac{x}{2}\right) \cdot \tan\frac{x}{2} \cdot \left(\frac{1}{2}\right) \cdot e^{-x/2}(-2) \, dx \right] + \frac{1}{2}\int e^{-x/2} \cdot \sec\frac{x}{2}\tan\frac{x}{2} \, dx \)
\( = e^{-x/2} \cdot \sec\frac{x}{2} - \frac{1}{2}\int e^{-x/2} \cdot \sec\frac{x}{2}\tan\frac{x}{2} \, dx + \frac{1}{2}\int e^{-x/2} \sec\frac{x}{2}\tan\frac{x}{2} \, dx \)
\( = e^{-x/2}\sec\left(\frac{x}{2}\right) + c \)      ans.

Type: \( \int \sqrt{\text{Quadratic}} \, dx \) and \( \int \text{Linear}\sqrt{\text{Quadratic}} \, dx \)

Perfect Square. Use Long Formula

Question. (a) \( I = \int \sqrt{(x - 3)(5 - x)} \, dx \)      (b) \( I = \int \sqrt{2x^2 + 3x + 4} \, dx \)      (c) \( I = \int \sqrt{3 - 2x - 2x^2} \, dx \)
Answer:
(a) \( I = \int \sqrt{(x - 3)(5 - x)} \, dx \)
\( = \int \sqrt{5x - x^2 - 15 + 3x} \, dx \)
\( = \int \sqrt{-x^2 + 8x - 15} \, dx \)
\( = \int \sqrt{-[x^2 - 8x + 15]} \, dx \)
\( = \int \sqrt{-[(x - 4)^2 - 1]} \, dx \)
\( = \int \sqrt{1^2 - (x - 4)^2} \, dx \)
\( = \frac{(x-4)}{2}\sqrt{1 - (x - 4)^2} + \frac{1}{2}\sin^{-1}\left(\frac{x-4}{1}\right) + c \)
\( = \frac{(x-4)}{2}\sqrt{(x - 3)(5 - x)} + \frac{1}{2}\sin^{-1}(x - 4) + c \)      ans.

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

Question. (a) \( I = \int (3x - 2)\sqrt{x^2 + x + 1} \, dx \)      (b) \( I = \int (4x + 1)\sqrt{x^2 - x - 2} \, dx \)
Answer:
(a) \( I = \int (3x - 2)\sqrt{x^2 + x + 1} \, dx \)
(take \( 2x + 1 \))
\( = 3\int \left(x - \frac{2}{3}\right)\sqrt{x^2 + x + 1} \, dx \)
\( = \frac{3}{2}\int \left(2x - \frac{4}{3}\right)\sqrt{x^2 + x + 1} \, dx \)
\( = \frac{3}{2}\int \left(2x - \frac{4}{3} + 1 - 1\right)\sqrt{x^2 + x + 1} \, dx \)
\( = \frac{3}{2}\int (2x + 1)\sqrt{x^2 + x + 1} \, dx - \frac{7}{2}\int \sqrt{x^2 + x + 1} \, dx \)
put \( x^2 + x + 1 = t \) in I
\( (2x + 1)dx = dt \)
\( = \frac{3}{2}\int \sqrt{t} \, dt - \frac{7}{2}\int \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + 1} \, dx \)
\( = \frac{3}{2} \times \frac{2}{3}(t)^{3/2} - \frac{7}{2}\int \sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dx \)
\( = (t)^{3/2} - \frac{7}{2} \left[ \frac{\left(x+\frac{1}{2}\right)}{2}\sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} + \frac{3}{8}\log\left| \left(x + \frac{1}{2}\right) + \sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \right| \right] \)
\( = (x^2 + x + 1)^{3/2} - \frac{7}{2}\left[ \frac{(2x+1)}{4}\sqrt{x^2 + x + 1} + \frac{3}{8}\log\left| \frac{2x+1}{2} + \sqrt{x^2 + x + 1} \right| \right] \)

(b) \( I = \int (4x + 1)\sqrt{x^2 - x - 2} \, dx \)
\( = 2\int \left(2x + \frac{1}{2}\right)\sqrt{x^2 - x - 2} \, dx \)
\( = 2\int \left(2x + \frac{1}{2} - 1 + 1\right)\sqrt{x^2 - x - 2} \, dx \)
\( = 2\int \left(2x - 1 + \frac{3}{2}\right)\sqrt{x^2 - x - 2} \, dx \)
\( = 2\int (2x - 1)\sqrt{x^2 - x - 2} \, dx + 3\int \sqrt{x^2 - x - 2} \, dx \)
put \( x^2 - x - 2 = t \)
\( (2x - 1)dx = dt \)
\( \dots I = 2\int \sqrt{t} \, dt + 3\int \sqrt{\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 2} \, dx \)
\( = 2 \times \frac{2}{3}(t)^{3/2} + 3\int \sqrt{\left(x - \frac{1}{2}\right)^2 - \left(\frac{3}{2}\right)^2} \, dx \)
\( = \frac{4}{3}(t)^{3/2} + 3\left[ \frac{2x-1}{4}\sqrt{x^2 - x - 2} - \frac{9}{8}\log\left| \frac{2x-1}{2} + \sqrt{x^2 - x - 2} \right| \right] \)      ans.

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