CBSE Class 12 Mathematics Integrals Assignment Set 01

Very Short Answer Type Questions (VSA)

Question. Write the antiderivative of \( \left(3\sqrt{x} + \frac{1}{\sqrt{x}}\right) \).
Answer: \( \int \left(3\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = 3 \int x^{1/2} dx + \int x^{-1/2} dx = 3 \cdot \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C = 2x\sqrt{x} + 2\sqrt{x} + C = 2\sqrt{x}(x+1) + C \)

Question. Evaluate: \( \int \cos^{-1}(\sin x) dx \)
Answer: \( \int \cos^{-1}(\sin x) dx = \int \cos^{-1}\left[\cos\left(\frac{\pi}{2}-x\right)\right] dx = \int \left(\frac{\pi}{2}-x\right) dx = \frac{\pi}{2}x - \frac{x^2}{2} + C \)

Question. Write the value of \( \int \frac{\sec^2 x}{\csc^2 x} dx \).
Answer: \( \int \frac{\sec^2 x}{\csc^2 x} dx = \int \frac{\sin^2 x}{\cos^2 x} dx = \int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C \)

Question. Write the value of \( \int \frac{2-3\sin x}{\cos^2 x} dx \).
Answer: \( \int \frac{2-3\sin x}{\cos^2 x} dx = \int \left(\frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x}\right) dx = \int (2\sec^2 x - 3\sec x \tan x) dx = 2\tan x - 3\sec x + C \)

Question. Evaluate: \( \int \frac{(\log x)^2}{x} dx \)
Answer: Let \( I = \int \frac{(\log x)^2}{x} dx \).
Put \( \log x = t \Rightarrow \frac{1}{x} dx = dt \).
Then, \( I = \int t^2 dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C \).

Question. Find: \( \int \frac{dx}{9 + 4x^2} \)
Answer: Let \( I = \int \frac{dx}{9 + 4x^2} = \frac{1}{4} \int \frac{dx}{x^2 + \left(\frac{3}{2}\right)^2} \).
Using \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \):
\( I = \frac{1}{4} \cdot \frac{1}{\frac{3}{2}} \tan^{-1}\left(\frac{x}{\frac{3}{2}}\right) + C = \frac{1}{6} \tan^{-1}\left(\frac{2x}{3}\right) + C \).

Question. Find: \( \int x^4 \log x dx \)
Answer: Let \( I = \int x^4 \log x dx \). Using integration by parts:
\( I = \log x \cdot \frac{x^5}{5} - \int \frac{1}{x} \cdot \frac{x^5}{5} dx = \frac{x^5}{5} \log x - \frac{1}{5} \int x^4 dx = \frac{x^5}{5} \log x - \frac{x^5}{25} + C \).

Question. If \( \int_0^a \frac{1}{4+x^2} dx = \frac{\pi}{8} \), find the value of \( a \).
Answer: We have \( \int_0^a \frac{1}{2^2 + x^2} dx = \frac{\pi}{8} \).
\( \Rightarrow \left[ \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right]_0^a = \frac{\pi}{8} \)
\( \Rightarrow \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{8} \)
\( \Rightarrow \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{4} \)
\( \Rightarrow \frac{a}{2} = \tan\left(\frac{\pi}{4}\right) = 1 \)
\( \Rightarrow a = 2 \).

Question. Write the value of \( \int_{0}^{1} \frac{e^x}{1 + e^{2x}} dx \).
Answer: Let \( I = \int_{0}^{1} \frac{e^x}{1+e^{2x}} dx \)
Put \( e^x = t \Rightarrow e^x dx = dt \)
Also, when \( x = 0 \Rightarrow t = e^0 = 1 \) and \( x = 1 \Rightarrow t = e^1 = e \)
\( \therefore I = \int_{1}^{e} \frac{dt}{1+t^2} = [\tan^{-1} t]_1^e = \tan^{-1} e - \tan^{-1} 1 = \tan^{-1}\left( \frac{e-1}{1+e} \right) \)

Question. Find the value of \( \int_{1}^{4} |x - 5| dx \).
Answer: Let \( I = \int_{1}^{4} |x - 5| dx \)
Since \( x \in [1, 4] \), \( |x - 5| = -(x - 5) \).
\( \therefore I = -\int_{1}^{4} (x - 5) dx = -\left[ \frac{x^2}{2} - 5x \right]_1^4 \)
\( = -\left[ \left(\frac{16}{2} - 20\right) - \left(\frac{1}{2} - 5\right) \right] \)
\( = -\left[ -12 - \left(-\frac{9}{2}\right) \right] = -\left[ -\frac{15}{2} \right] = \frac{15}{2} \)

Short Answer Type Questions (SA-I)

Question. Find : \( \int \sqrt{1 - \sin 2x} dx, \, \frac{\pi}{4} < x < \frac{\pi}{2} \)
Answer: Let \( I = \int \sqrt{1 - \sin 2x} dx \)
\( = \int \sqrt{\cos^2 x + \sin^2 x - 2\sin x \cos x} dx \)
\( = \pm \int (\cos x - \sin x) dx \)
Since \( \frac{\pi}{4} < x < \frac{\pi}{2} \), we get:
\( I = \int (\sin x - \cos x) dx = -(\cos x + \sin x) + C \)

Question. Evaluate : \( \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} dx \)
Answer: Let \( I = \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} dx \)
\( = \int \frac{\cos^2 x - \sin^2 x + 2\sin^2 x}{\cos^2 x} dx \)
\( = \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x} dx = \int \frac{1}{\cos^2 x} dx \)
\( = \int \sec^2 x \, dx = \tan x + C \)

Question. Find : \( \int \frac{dx}{\sqrt{5 - 4x - 2x^2}} \)
Answer: Let \( I = \int \frac{dx}{\sqrt{5 - 4x - 2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{7}{2} - 1 - 2x - x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2 - (x+1)^2}} \)
\( = \frac{1}{\sqrt{2}} \sin^{-1}\left( \frac{x+1}{\sqrt{7/2}} \right) + C = \frac{1}{\sqrt{2}} \sin^{-1}\left( \sqrt{\frac{2}{7}}(x+1) \right) + C \)

Question. Find : \( \int \frac{dx}{x^2 + 4x + 8} \)
Answer: We have, \( \int \frac{dx}{x^2 + 4x + 8} = \int \frac{dx}{x^2 + 4x + 4 + 4} \)
\( = \int \frac{dx}{(x+2)^2 + (2)^2} = \frac{1}{2} \tan^{-1}\left(\frac{x+2}{2}\right) + C \)

Question. Find \( \int \frac{x+1}{(x+2)(x+3)} dx \).
Answer: Let \( I = \int \frac{x+1}{(x+2)(x+3)} dx \)
Also let, \( \frac{x+1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} \)
\( \Rightarrow x + 1 = A(x+3) + B(x+2) \quad \dots(i) \)
Putting \( x = -3 \) in \( (i) \), we get:
\( -B = -3 + 1 = -2 \Rightarrow B = 2 \)
Putting \( x = -2 \) in \( (i) \), we get:
\( A = -2 + 1 = -1 \)
\( \therefore I = \int \frac{-1}{x+2} dx + \int \frac{2}{x+3} dx \)
\( = -\log(x+2) + 2\log(x+3) + C \)

Question. Find : \( \int \sin^{-1}(2x) dx \)
Answer: Let \( I = \int \sin^{-1}(2x) dx = \int 1 \cdot \sin^{-1}(2x) dx \)
Integrating by parts, we get:
\( = \sin^{-1}(2x) \cdot x - \int \left( \frac{1}{\sqrt{1 - 4x^2}} \cdot (2) \cdot x \right) dx \)
\( = x \sin^{-1}(2x) - \int \frac{2x}{\sqrt{1 - 4x^2}} dx \)
\( = x \sin^{-1}(2x) + \frac{1}{4} \int \frac{dt}{\sqrt{t}} \quad \text{(Putting } 1 - 4x^2 = t \Rightarrow -8xdx = dt\text{)} \)
\( = x \sin^{-1}(2x) + \frac{2}{4} (t)^{1/2} + C \)
\( = x \sin^{-1}(2x) + \frac{1}{2}\sqrt{1 - 4x^2} + C \)

Question. Find : \( \int x \cdot \tan^{-1} x \, dx \)
Answer: Let \( I = \int x \cdot \tan^{-1} x \, dx \)
Integrating by parts, we get:
\( I = \tan^{-1} x \int x dx - \int \left\{ \left( \frac{d}{dx}(\tan^{-1} x) \right) \int x dx \right\} dx \)
\( = (\tan^{-1} x) \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx \)
\( = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int dx + \frac{1}{2} \int \frac{1}{1+x^2} dx \)
\( = \frac{x^2 \tan^{-1} x}{2} - \frac{x}{2} + \frac{1}{2}\tan^{-1} x + C \)
\( = \frac{1}{2}(1 + x^2)\tan^{-1} x - \frac{x}{2} + C \)

Question. Evaluate \( \int_{1}^{2} \left[ \frac{1}{x} - \frac{1}{2x^2} \right] e^{2x} dx \).
Answer: Let \( I = \int_{1}^{2} \left[ \frac{1}{x} - \frac{1}{2x^2} \right] e^{2x} dx \)
Putting \( 2x = y \Rightarrow 2dx = dy \)
As \( x \to 1 \Rightarrow y \to 2 \) and \( x \to 2 \Rightarrow y \to 4 \)
\( \therefore I = \frac{1}{2} \int_{2}^{4} \left[ \frac{2}{y} - \frac{2}{y^2} \right] e^y dy = \int_{2}^{4} \left[ \frac{1}{y} - \frac{1}{y^2} \right] e^y dy \)
\( = \left[ \frac{e^y}{y} \right]_2^4 = \frac{e^4}{4} - \frac{e^2}{2} = \frac{e^2}{2}\left( \frac{e^2}{2} - 1 \right) \)

Question. Find the value of \( \int_{0}^{1} \tan^{-1}\left( \frac{1-2x}{1+x-x^2} \right) dx \).
Answer: Let \( I = \int_{0}^{1} \tan^{-1}\left( \frac{1-2x}{1+x-x^2} \right) dx \)
\( = \int_{0}^{1} \tan^{-1}\left( \frac{(1-x) - x}{1 + x(1-x)} \right) dx \)
\( I = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1} x] dx \quad \dots(i) \)
\( I = \int_{0}^{1} [\tan^{-1} x - \tan^{-1}(1-x)] dx \quad \dots(ii) \quad \left[ \text{Using property, } \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \right] \)
Adding \( (i) \) and \( (ii) \), we get:
\( 2I = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1} x + \tan^{-1} x - \tan^{-1}(1-x)] dx = 0 \)
\( \Rightarrow I = 0 \)

Question. Find: \( \int_{-\pi/4}^{0} \frac{1+\tan x}{1-\tan x} dx \)
Answer: Let \( I = \int_{-\pi/4}^{0} \frac{1+\tan x}{1-\tan x} dx = \int_{-\pi/4}^{0} \frac{\cos x + \sin x}{\cos x - \sin x} dx \)
Put \( \cos x - \sin x = t \Rightarrow -(\sin x + \cos x) dx = dt \)
When \( x = 0, \, t = 1 \), and when \( x = -\frac{\pi}{4}, \, t = \sqrt{2} \)
\( \therefore I = \int_{\sqrt{2}}^{1} \frac{-dt}{t} = \int_{1}^{\sqrt{2}} \frac{dt}{t} = [\log |t|]_1^{\sqrt{2}} \)
\( = \log\sqrt{2} - \log 1 = \frac{1}{2}\log 2 \)

Short Answer Type Questions (SA-II)

Question. Evaluate : \( \int \frac{\sin(x - a)}{\sin(x + a)} dx \)
Answer: Let \( I = \int \frac{\sin(x - a)}{\sin(x + a)} dx = \int \frac{\sin(x + a - 2a)}{\sin(x + a)} dx \)
\( = \int \left[ \frac{\sin(x+a)\cos 2a - \cos(x+a)\sin 2a}{\sin(x+a)} \right] dx \)
\( \Rightarrow I = \cos 2a \int dx - \sin 2a \int \frac{\cos(x+a)}{\sin(x+a)} dx \)
Put \( \sin(x+a) = t \Rightarrow \cos(x+a)dx = dt \)
\( \Rightarrow I = \cos 2a \int dx - \sin 2a \int \frac{dt}{t} \)
\( = x \cos 2a - \sin 2a \log|\sin(x+a)| + C \)

Question. Evaluate : \( \int \sin x \sin 2x \sin 3x dx \)
Answer: \( \int \sin 3x \sin 2x \sin x dx \)
\( = \frac{1}{2} \int (\cos 2x - \cos 4x) \sin 2x dx \)
\( = \frac{1}{2} \int \sin 2x \cos 2x dx - \frac{1}{2} \int \cos 4x \sin 2x dx \)
\( = \frac{1}{4} \int \sin 4x dx - \frac{1}{4} \int (\sin 6x - \sin 2x) dx \)
\( = \frac{1}{4} \int \sin 4x dx - \frac{1}{4} \int \sin 6x dx + \frac{1}{4} \int \sin 2x dx \)
\( = \frac{1}{4} \left[ \frac{-\cos 4x}{4} \right] - \frac{1}{4} \left[ \frac{-\cos 6x}{6} \right] + \frac{1}{4} \left[ \frac{-\cos 2x}{2} \right] + C \)
\( = \frac{1}{4} \left[ \frac{\cos 6x}{6} - \frac{\cos 4x}{4} - \frac{\cos 2x}{2} \right] + C \)

Question. Find \( \int (\sqrt{\tan x} + \sqrt{\cot x}) dx \).
Answer: Let \( I = \int (\sqrt{\tan x} + \sqrt{\cot x}) dx \)
\( = \int \left( \frac{\sin x}{\sqrt{\cos x}} + \frac{\cos x}{\sqrt{\sin x}} \right) dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx \)
\( = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (1 - 2\sin x \cos x)}} dx \)
\( = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} dx \)
Put \( \sin x - \cos x = t \Rightarrow (\cos x + \sin x)dx = dt \)
\( \therefore I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2}\sin^{-1} t + C \)
\( = \sqrt{2}\sin^{-1}(\sin x - \cos x) + C \)

Question. Evaluate : \( \int \frac{x+2}{\sqrt{x^2+5x+6}} dx \)
Answer: Let \( I = \int \frac{x+2}{\sqrt{x^2+5x+6}} dx = \int \frac{\frac{1}{2}(2x+5) - \frac{1}{2}}{\sqrt{x^2+5x+6}} dx \)
\( = \frac{1}{2} \int (x^2+5x+6)^{-1/2}(2x+5) dx - \frac{1}{2} \int \frac{dx}{\sqrt{x^2+5x+6}} \)
Put \( x^2+5x+6 = t \Rightarrow (2x+5)dx = dt \)
\( \Rightarrow I = \frac{1}{2} \int t^{-1/2} dt - \frac{1}{2} \int \frac{dx}{\sqrt{\left(x+\frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} + C \)
\( = t^{1/2} - \frac{1}{2} \log\left| \left(x+\frac{5}{2}\right) + \sqrt{\left(x+\frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C \)
\( = \sqrt{x^2+5x+6} - \frac{1}{2}\log\left| x + \frac{5}{2} + \sqrt{x^2+5x+6} \right| + C \)

Question. Evaluate : \( \int \frac{5x+3}{\sqrt{x^2+4x+10}} dx \)
Answer: Let \( I = \int \frac{5x+3}{\sqrt{x^2+4x+10}} dx = \int \frac{\frac{5}{2}(2x+4) - 7}{\sqrt{x^2+4x+10}} dx \)
\( = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx - 7 \int \frac{dx}{\sqrt{x^2+4x+10}} \)
\( = I_1 + I_2 \text{ (say)} \quad \dots(1) \)
where \( I_1 = \frac{5}{2} \int \frac{2x+4}{\sqrt{x^2+4x+10}} dx \)
Put \( x^2 + 4x + 10 = t \Rightarrow (2x+4)dx = dt \)
\( \therefore I_1 = \frac{5}{2} \int t^{-1/2} dt = 5\sqrt{t} = 5\sqrt{x^2+4x+10} + C_1 \quad \dots(2) \)
and \( I_2 = -7 \int \frac{dx}{\sqrt{x^2+4x+10}} = -7 \int \frac{dx}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} \)
\( = -7\log|x+2+\sqrt{x^2+4x+10}| + C_2 \quad \dots(3) \)
From \( (1), (2) \) and \( (3) \), we get:
\( I = 5\sqrt{x^2+4x+10} - 7\log|x+2+\sqrt{x^2+4x+10}| + C \), where \( C = C_1 + C_2 \)

Question. Evaluate : \( \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx \)
Answer: Let \( I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx = \int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2x - 1}} dx \)
\( = \int \frac{\sin x - \cos x}{\sqrt{\sin^2 x + \cos^2 x + 2\sin x \cos x - 1}} dx \)
\( = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} dx \)
Put \( \sin x + \cos x = t \Rightarrow (\cos x - \sin x)dx = dt \)
\( \therefore I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\log|t + \sqrt{t^2 - 1}| + C \)
\( = -\log|\sin x + \cos x + \sqrt{\sin 2x}| + C \)

Question. Find : \( \int \frac{(x^2 + 1)(x^2 + 4)}{(x^2 + 3)(x^2 - 5)} dx \)
Answer: Let \( I = \int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)} dx \)
Let \( x^2 = t \)
\( \therefore \frac{(t+1)(t+4)}{(t+3)(t-5)} = \frac{t^2+5t+4}{t^2-2t-15} = 1 + \frac{7t+19}{(t+3)(t-5)} \)
Let \( \frac{7t+19}{(t+3)(t-5)} = \frac{A}{t+3} + \frac{B}{t-5} \)
\( \Rightarrow 7t+19 = A(t-5) + B(t+3) \)
Putting \( t = 5 \), we get \( B = \frac{27}{4} \)
Putting \( t = -3 \), we get \( A = \frac{1}{4} \)
\( \therefore \frac{t^2+5t+4}{(t+3)(t-5)} = 1 + \frac{1}{4(t+3)} + \frac{27}{4(t-5)} \)
\( \Rightarrow I = \int dx + \frac{1}{4} \int \frac{1}{x^2+3} dx + \frac{27}{4} \int \frac{1}{x^2-5} dx \)
\( = x + \frac{1}{4\sqrt{3}}\tan^{-1}\left( \frac{x}{\sqrt{3}} \right) + \frac{27}{8\sqrt{5}}\log\left| \frac{x-\sqrt{5}}{x+\sqrt{5}} \right| + C \)

Question. Find : \( \int \frac{x}{(x^2+1)(x-1)} dx \)
Answer: Let \( I = \int \frac{x}{(x^2+1)(x-1)} dx \)
Let \( \frac{x}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1} \quad \dots(1) \)
\( \Rightarrow x = (Ax+B)(x-1) + C(x^2+1) \quad \dots(2) \)
Comparing coefficients of \( x^2, \, x \) and constant terms, we get:
\( A + C = 0; \quad B - A = 1; \quad -B + C = 0 \)
Solving these, we get: \( A = -\frac{1}{2}, \quad C = \frac{1}{2}, \quad B = \frac{1}{2} \)
\( \therefore \frac{x}{(x^2+1)(x-1)} = -\frac{1}{2}\left(\frac{x-1}{x^2+1}\right) + \frac{1}{2}\left(\frac{1}{x-1}\right) \)
\( = -\frac{1}{4}\left(\frac{2x}{x^2+1}\right) + \frac{1}{2}\left(\frac{1}{x^2+1}\right) + \frac{1}{2}\left(\frac{1}{x-1}\right) \)
\( \therefore I = -\frac{1}{4}\log|x^2+1| + \frac{1}{2}\tan^{-1} x + \frac{1}{2}\log|x-1| + C \)

Question. Evaluate : \( \int e^{2x} \cdot \sin(3x + 1) dx \)
Answer: Let \( I = \int e^{2x} \cdot \sin(3x + 1) dx \)
Integrating by parts, we get:
\( = e^{2x} \left[ \frac{-\cos(3x+1)}{3} \right] - \int 2e^{2x} \cdot \left[ \frac{-\cos(3x+1)}{3} \right] dx \)
\( = -\frac{e^{2x}\cos(3x+1)}{3} + \frac{2}{3} \int e^{2x}\cos(3x+1) dx \)
\( = -\frac{e^{2x}\cos(3x+1)}{3} + \frac{2}{3} \left[ e^{2x}\frac{\sin(3x+1)}{3} - \int 2e^{2x}\frac{\sin(3x+1)}{3} dx \right] \)
\( = -\frac{e^{2x}\cos(3x+1)}{3} + \frac{2}{9} e^{2x}\sin(3x+1) - \frac{4}{9}I + C_1 \)
\( \therefore I + \frac{4}{9}I = -\frac{e^{2x}\cos(3x+1)}{3} + \frac{2}{9} e^{2x}\sin(3x+1) + C_1 \)
\( \Rightarrow \frac{13}{9}I = \frac{e^{2x}}{9} [2\sin(3x+1) - 3\cos(3x+1)] + C_1 \)
\( \Rightarrow I = \frac{e^{2x}}{13} [2\sin(3x+1) - 3\cos(3x+1)] + C \)

Question. Evaluate : \( \int \frac{x\cos^{-1} x}{\sqrt{1-x^2}} dx \)
Answer: Let \( I = \int \frac{x\cos^{-1} x}{\sqrt{1-x^2}} dx \)
Put \( \cos^{-1} x = \theta \Rightarrow x = \cos\theta \Rightarrow dx = -\sin\theta d\theta \)
\( \Rightarrow I = \int \frac{\theta \cos\theta}{\sqrt{1-\cos^2\theta}} (-\sin\theta d\theta) \Rightarrow I = -\int \theta \cos\theta d\theta \)
Integrating by parts:
\( -I = \theta \sin\theta - \int \sin\theta d\theta \)
\( -I = \theta \sin\theta + \cos\theta + C_1 \)
\( \Rightarrow I = -[\theta\sin\theta + \cos\theta] + C \)
\( \therefore I = -\left[ \sqrt{1-x^2}\cos^{-1} x + x \right] + C \)

Question. Evaluate : \( \int_{0}^{\pi/2} x^2 \sin x \, dx \)
Answer: Let \( I = \int_{0}^{\pi/2} x^2 \sin x dx \)
Integrating by parts, we get:
\( I = [x^2(-\cos x)]_0^{\pi/2} - \int_{0}^{\pi/2} 2x(-\cos x) dx \)
\( = 0 + 2 \int_{0}^{\pi/2} x \cos x dx \)
Again integrating by parts:
\( I = 2 \left[ [x\sin x]_0^{\pi/2} - \int_{0}^{\pi/2} 1 \cdot \sin x dx \right] \)
\( = 2 \left[ \frac{\pi}{2} - [-\cos x]_0^{\pi/2} \right] = 2 \left[ \frac{\pi}{2} + (0 - 1) \right] = \pi - 2 \)

Question. Evaluate \( \int_{0}^{\pi} e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx \).
Answer: Let \( I = \int_{0}^{\pi} e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx \)
Put \( \frac{\pi}{4} + x = t \Rightarrow x = t - \frac{\pi}{4} \Rightarrow dx = dt \)
When \( x = 0, \, t = \frac{\pi}{4} \) and when \( x = \pi, \, t = \frac{5\pi}{4} \)
\( \therefore I = \int_{\pi/4}^{5\pi/4} e^{2(t-\pi/4)}\sin t dt = e^{-\pi/2} \int_{\pi/4}^{5\pi/4} e^{2t}\sin t dt \)
Using integration by parts twice, we get:
\( e^{\pi/2} I = \left[ \frac{e^{2t}}{5}(2\sin t - \cos t) \right]_{\pi/4}^{5\pi/4} \)
\( \Rightarrow e^{\pi/2} I = \frac{e^{5\pi/2}}{5}\left(2\sin\frac{5\pi}{4} - \cos\frac{5\pi}{4}\right) - \frac{e^{\pi/2}}{5}\left(2\sin\frac{\pi}{4} - \cos\frac{\pi}{4}\right) \)
\( \Rightarrow e^{\pi/2} I = \frac{e^{5\pi/2}}{5}\left( -\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \frac{e^{\pi/2}}{5}\left( \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \)
\( \Rightarrow e^{\pi/2} I = -\frac{e^{5\pi/2}}{5\sqrt{2}} - \frac{e^{\pi/2}}{5\sqrt{2}} \)
\( \Rightarrow I = -\frac{1}{5\sqrt{2}}(e^{2\pi} + 1) \)

Question. Evaluate : \( \int_{0}^{\pi} \frac{x}{1 + \sin\alpha \sin x} dx \)
Answer: Let \( I = \int_{0}^{\pi} \frac{x}{1 + \sin\alpha \sin x} dx \)
\( \Rightarrow I = \int_{0}^{\pi} \frac{\pi-x}{1 + \sin\alpha \sin(\pi-x)} dx = \int_{0}^{\pi} \frac{\pi-x}{1 + \sin\alpha \sin x} dx \)
Adding the equations:
\( 2I = \pi \int_{0}^{\pi} \frac{1}{1 + \sin\alpha \sin x} dx \)
Using half-angle formula \( \sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)} \):
\( 2I = \pi \int_{0}^{\pi} \frac{\sec^2(x/2)}{1 + \tan^2(x/2) + 2\sin\alpha\tan(x/2)} dx \)
Put \( \tan(x/2) = t \Rightarrow \sec^2(x/2)dx = 2dt \). At limits, \( x \to 0 \Rightarrow t \to 0 \) and \( x \to \pi \Rightarrow t \to \infty \).
\( \Rightarrow I = \pi \int_{0}^{\infty} \frac{dt}{(t+\sin\alpha)^2 + \cos^2\alpha} \)
\( = \frac{\pi}{\cos\alpha} \left[ \tan^{-1}\left(\frac{t+\sin\alpha}{\cos\alpha}\right) \right]_0^{\infty} \)
\( = \frac{\pi}{\cos\alpha} \left[ \frac{\pi}{2} - \tan^{-1}(\tan\alpha) \right] = \frac{\pi}{\cos\alpha}\left( \frac{\pi}{2} - \alpha \right) \)

Question. Evaluate : \( \int_{1}^{4} [|x - 1| + |x - 2| + |x - 4|] dx \)
Answer: Let \( f(x) = |x - 1| + |x - 2| + |x - 4| \)
We have critical points \( x = 1, 2, 4 \).
\( f(x) = \begin{cases} (x-1) - (x-2) - (x-4) = -x + 5, & \text{if } 1 \le x < 2 \\ (x-1) + (x-2) - (x-4) = x + 1, & \text{if } 2 \le x \le 4 \end{cases} \)
\( \therefore I = \int_{1}^{2} (-x + 5) dx + \int_{2}^{4} (x + 1) dx \)
\( = \left[ -\frac{x^2}{2} + 5x \right]_1^2 + \left[ \frac{x^2}{2} + x \right]_2^4 \)
\( = \left( -2 + 10 \right) - \left( -\frac{1}{2} + 5 \right) + (8 + 4) - (2 + 2) \)
\( = 8 - \frac{9}{2} + 12 - 4 = 16 - \frac{9}{2} = \frac{23}{2} \)

Question. Using properties of definite integrals, evaluate the following : \( \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx \)
Answer: Let \( I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \dots(1) \)
Using property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \):
\( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\pi/2-x)}}{\sqrt{\cos(\pi/2-x)} + \sqrt{\sin(\pi/2-x)}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots(2) \)
Adding \( (1) \) and \( (2) \), we get:
\( 2I = \int_{\pi/6}^{\pi/3} 1 \cdot dx = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)
\( \Rightarrow I = \frac{\pi}{12} \pmb \)

Long Answer Type Questions (LA)

Question. Evaluate : \( \int \frac{6x+7}{\sqrt{(x-5)(x-4)}} dx \)
Answer: Let \( I = \int \frac{6x+7}{\sqrt{x^2 - 9x + 20}} dx \)
Let \( 6x + 7 = A \frac{d}{dx}(x^2 - 9x + 20) + B \Rightarrow 6x + 7 = A(2x - 9) + B \)
Comparing coefficients, we get:
\( 2A = 6 \Rightarrow A = 3 \)
\( -9A + B = 7 \Rightarrow -27 + B = 7 \Rightarrow B = 34 \)
\( \therefore I = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34 \int \frac{dx}{\sqrt{x^2-9x+20}} \)
\( = 3 (2\sqrt{x^2-9x+20}) + 34 \int \frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)
\( = 6\sqrt{x^2-9x+20} + 34\log\left| \left(x-\frac{9}{2}\right) + \sqrt{x^2-9x+20} \right| + C \)

Question. Evaluate : \( \int \frac{x^2+1}{(x-1)^2(x+3)} dx \)
Answer: Let \( \frac{x^2+1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3} \)
\( \Rightarrow x^2 + 1 = A(x-1)(x+3) + B(x+3) + C(x-1)^2 \)
Putting \( x = 1 \Rightarrow 2 = 4B \Rightarrow B = \frac{1}{2} \)
Putting \( x = -3 \Rightarrow 10 = 16C \Rightarrow C = \frac{5}{8} \)
Putting \( x = 0 \Rightarrow 1 = -3A + 3B + C \Rightarrow A = \frac{3}{8} \)
Integrating both sides:
\( I = \frac{3}{8}\log|x-1| - \frac{1}{2(x-1)} + \frac{5}{8}\log|x+3| + C \)

Question. Find : \( \int \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} dx, \quad x \in [0,1] \)
Answer: We know that \( \sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2} \)
\( \therefore I = \int \frac{\frac{\pi}{2} - 2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}} dx = \int dx - \frac{4}{\pi} \int \cos^{-1}\sqrt{x} \, dx \)
\( = x - \frac{4}{\pi} \left[ x \cos^{-1}\sqrt{x} - \int x \left( -\frac{1}{\sqrt{1-x}} \frac{1}{2\sqrt{x}} \right) dx \right] \)
Put \( x = \sin^2\theta \Rightarrow dx = 2\sin\theta\cos\theta d\theta \):
\( = x - \frac{4}{\pi}x\cos^{-1}\sqrt{x} - \frac{4}{\pi} \left[ \theta - \sin\theta\cos\theta \right] + C \)
\( = x - \frac{4}{\pi}x\cos^{-1}\sqrt{x} - \frac{4}{\pi} \left[ \sin^{-1}\sqrt{x} - \sqrt{x}\sqrt{1-x} \right] + C \)

Question. Prove that \( \int_{0}^{\pi/4} (\sqrt{\tan x} + \sqrt{\cot x}) dx = \sqrt{2} \cdot \frac{\pi}{2} \)
Answer: \( \text{L.H.S.} = \int_{0}^{\pi/4} (\sqrt{\tan x} + \sqrt{\cot x}) dx = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{\sqrt{\sin x\cos x}} dx \)
\( = \sqrt{2} \int_{0}^{\pi/4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} dx \)
Put \( \sin x - \cos x = t \Rightarrow (\cos x + \sin x)dx = dt \)
At \( x = 0, \, t = -1 \); at \( x = \pi/4, \, t = 0 \).
\( \therefore \text{L.H.S.} = \sqrt{2} \int_{-1}^{0} \frac{dt}{\sqrt{1-t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^{0} \)
\( = \sqrt{2} [\sin^{-1} 0 - \sin^{-1}(-1)] = \sqrt{2} \left[ 0 - \left(-\frac{\pi}{2}\right) \right] = \sqrt{2} \cdot \frac{\pi}{2} = \text{R.H.S.} \)

Question. Evaluate: \( \int_{0}^{\pi} \frac{x dx}{a^2 \cos^2 x + b^2 \sin^2 x} \)
Answer: Let \( I = \int_{0}^{\pi} \frac{x dx}{a^2 \cos^2 x + b^2 \sin^2 x} \quad \dots(1) \)
Using property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \):
\( I = \int_{0}^{\pi} \frac{(\pi-x) dx}{a^2 \cos^2 x + b^2 \sin^2 x} \quad \dots(2) \)
Adding \( (1) \) and \( (2) \):
\( 2I = \pi \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \)
Using symmetry about \( \pi/2 \):
\( 2I = 2\pi \int_{0}^{\pi/2} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x} \Rightarrow I = \pi \int_{0}^{\pi/2} \frac{\sec^2 x dx}{a^2 + b^2 \tan^2 x} \)
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \). At limits, \( x \to 0 \Rightarrow t \to 0 \); \( x \to \pi/2 \Rightarrow t \to \infty \).
\( I = \pi \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{\pi}{b^2} \int_{0}^{\infty} \frac{dt}{\left(\frac{a}{b}\right)^2 + t^2} \)
\( = \frac{\pi}{b^2} \cdot \frac{b}{a} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_0^{\infty} = \frac{\pi}{ab} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi^2}{2ab} \)

Please click the below link to access CBSE Class 12 Mathematics Integrals Assignment Set A