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, \( x \to 0 \Rightarrow t \to 0 \); \( x \to \pi/2 \Rightarrow t \to \infty \).
\( I = \pi \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{\pi}{b^2} \int_{0}^{\infty} \frac{dt}{\left(\frac{a}{b}\right)^2 + t^2} \)
\( = \frac{\pi}{b^2} \cdot \frac{b}{a} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_0^{\infty} = \frac{\pi}{ab} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi^2}{2ab} \)

Evaluate

Question 1. \(\int \tan x \tan 2x \tan 3x \, dx\)
Answer: We know that \(\tan 3x = \tan(2x + x)\).
Using the formula for \(\tan(A + B)\), we can write:
\(\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)
\(\tan 3x(1 - \tan 2x \tan x) = \tan 2x + \tan x\)
\(\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x\)
Rearranging the terms, we get:
\(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Now, integrating both sides with respect to \(x\):
\(\int \tan x \tan 2x \tan 3x \, dx = \int (\tan 3x - \tan 2x - \tan x) \, dx\)
\(= \frac{1}{3} \ln|\sec 3x| - \frac{1}{2} \ln|\sec 2x| - \ln|\sec x| + C\)
\(= -\frac{1}{3} \ln|\cos 3x| + \frac{1}{2} \ln|\cos 2x| + \ln|\cos x| + C\)
In simple words: Express the product of three tangents using the angle sum property of tangents, then integrate each term individually.

Exam Tip: Remember the trigonometric identity connecting \(\tan 3x\), \(\tan 2x\), and \(\tan x\) to simplify the integrand into standard integrable forms.

Question 2. \(\int \frac{\cos x}{(\cos \frac{x}{2} + \sin \frac{x}{2})^3} \, dx\)
Answer: Let the given integral be \(I\).
Using double angle formulas, \(\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})\).
Substitute this into the integral:
\(I = \int \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2})^3} \, dx\)
\(I = \int \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} \, dx\)
Put \(t = \cos \frac{x}{2} + \sin \frac{x}{2}\). Then, \(dt = \frac{1}{2} (-\sin \frac{x}{2} + \cos \frac{x}{2}) \, dx\), which implies \((\cos \frac{x}{2} - \sin \frac{x}{2}) \, dx = 2 \, dt\).
Substituting into the integral:
\(I = \int \frac{2}{t^2} \, dt = 2 \int t^{-2} \, dt = 2 \left( \frac{t^{-1}}{-1} \right) + C = -\frac{2}{t} + C\)
Substituting back the value of \(t\):
\(I = -\frac{2}{\cos \frac{x}{2} + \sin \frac{x}{2}} + C\)
In simple words: Expand \(\cos x\) into half-angles, cancel out common factors in the numerator and denominator, and use substitution.

Exam Tip: Converting double angles to half-angles is a powerful technique when denominators contain half-angle sums or differences.

Question 3. \(\int \sin^{-1} \sqrt{\frac{x}{a+x}} \, dx\)
Answer: Put \(x = a \tan^2 \theta\). Then \(dx = 2a \tan \theta \sec^2 \theta \, d\theta\).
Also, \(\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} = \sqrt{\frac{a \tan^2 \theta}{a \sec^2 \theta}} = \sin \theta\).
Therefore, \(\sin^{-1} \sqrt{\frac{x}{a+x}} = \sin^{-1}(\sin \theta) = \theta\).
The integral becomes:
\(I = \int \theta \cdot 2a \tan \theta \sec^2 \theta \, d\theta = 2a \int \theta (\tan \theta \sec^2 \theta) \, d\theta\)
Using integration by parts, taking \(\theta\) as the first function and \(\tan \theta \sec^2 \theta\) as the second function:
\(I = 2a \left[ \theta \int \tan \theta \sec^2 \theta \, d\theta - \int \left( \frac{d}{d\theta}(\theta) \int \tan \theta \sec^2 \theta \, d\theta \right) d\theta \right]\)
Since \(\int \tan \theta \sec^2 \theta \, d\theta = \frac{\tan^2 \theta}{2}\):
\(I = 2a \left[ \theta \cdot \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} \, d\theta \right]\)
\(I = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) \, d\theta\)
\(I = a \theta \tan^2 \theta - a(\tan \theta - \theta) + C\)
Substitute back \(\theta = \tan^{-1}\sqrt{\frac{x}{a}}\) and \(\tan^2 \theta = \frac{x}{a}\):
\(I = a \left( \tan^{-1}\sqrt{\frac{x}{a}} \right) \left( \frac{x}{a} \right) - a \left( \sqrt{\frac{x}{a}} - \tan^{-1}\sqrt{\frac{x}{a}} \right) + C\)
\(= x \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{ax} + a \tan^{-1}\sqrt{\frac{x}{a}} + C\)
In simple words: Substitute \(x = a \tan^2 \theta\) to simplify the inverse trigonometric function, then apply integration by parts.

Exam Tip: Watch out for proper substitution in inverse trigonometric integrands; trigonometric substitutions often turn complex radicals into simple polynomials.

Question 4. \(\int e^x \sin^2 x \, dx\)
Answer: Using the trigonometric identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\):
\(\int e^x \sin^2 x \, dx = \int e^x \left( \frac{1 - \cos 2x}{2} \right) dx\)
\(= \frac{1}{2} \int e^x \, dx - \frac{1}{2} \int e^x \cos 2x \, dx\)
The first integral is simply \(\frac{1}{2} e^x\).
For the second integral, let \(I_1 = \int e^x \cos 2x \, dx\). Using the standard formula \(\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx))\):
\(I_1 = \frac{e^x}{1^2 + 2^2} (\cos 2x + 2 \sin 2x) = \frac{e^x}{5} (\cos 2x + 2 \sin 2x)\)
Combining both parts:
\(\int e^x \sin^2 x \, dx = \frac{1}{2} e^x - \frac{1}{2} \left[ \frac{e^x}{5} (\cos 2x + 2 \sin 2x) \right] + C\)
\(= \frac{1}{2} e^x - \frac{e^x}{10} (\cos 2x + 2 \sin 2x) + C\)
In simple words: Convert \(\sin^2 x\) into a linear cosine term using multiple angle formulas, then integrate using standard exponential-trigonometric integration rules.

Exam Tip: Integrals containing products of exponentials and trigonometric functions can be evaluated quickly using standard reduction formulas or integration by parts twice.

Question 5. \(\int \frac{\sin x}{\sin 3x} \, dx\)
Answer: Using the identity \(\sin 3x = 3 \sin x - 4 \sin^3 x = \sin x (3 - 4 \sin^2 x)\), the integral becomes:
\(\int \frac{\sin x}{\sin x (3 - 4 \sin^2 x)} \, dx = \int \frac{1}{3 - 4 \sin^2 x} \, dx\)
Divide numerator and denominator by \(\cos^2 x\):
\(\int \frac{\sec^2 x}{3 \sec^2 x - 4 \tan^2 x} \, dx = \int \frac{\sec^2 x}{3(1 + \tan^2 x) - 4 \tan^2 x} \, dx\)
\(= \int \frac{\sec^2 x}{3 + 3 \tan^2 x - 4 \tan^2 x} \, dx = \int \frac{\sec^2 x}{3 - \tan^2 x} \, dx\)
Let \(\tan x = t\), then \(\sec^2 x \, dx = dt\):
\(\int \frac{dt}{3 - t^2} = \int \frac{dt}{(\sqrt{3})^2 - t^2}\)
Using the standard formula \(\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln\left| \frac{a+x}{a-x} \right|\):
\(= \frac{1}{2\sqrt{3}} \ln\left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| + C\)
Substitute back \(t = \tan x\):
\(= \frac{1}{2\sqrt{3}} \ln\left| \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} \right| + C\)
In simple words: Expand \(\sin 3x\), cancel \(\sin x\), convert the expression into secants and tangents, and use standard algebraic substitution.

Exam Tip: Whenever powers of sine and cosine appear in denominators, dividing by \(\cos^2 x\) to form terms of \(\tan x\) and \(\sec^2 x\) is a standard strategy.

Question 6. \(\int \frac{\sin x}{\sin 4x} \, dx\)
Answer: Write \(\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x\).
Substitute into the integral:
\(\int \frac{\sin x}{4 \sin x \cos x \cos 2x} \, dx = \frac{1}{4} \int \frac{dx}{\cos x \cos 2x}\)
Multiply numerator and denominator by \(\cos x\):
\(= \frac{1}{4} \int \frac{\cos x}{\cos^2 x \cos 2x} \, dx = \frac{1}{4} \int \frac{\cos x}{(1 - \sin^2 x)(1 - 2 \sin^2 x)} \, dx\)
Let \(\sin x = t\), then \(\cos x \, dx = dt\):
\(= \frac{1}{4} \int \frac{dt}{(1 - t^2)(1 - 2t^2)}\)
Using partial fractions:
\(\frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{2}{1 - 2t^2} - \frac{1}{1 - t^2}\)
Integrating each term separately:
\(\frac{1}{4} \int \left( \frac{2}{1 - 2t^2} - \frac{1}{1 - t^2} \right) dt = \frac{1}{4} \left[ \frac{2}{2\sqrt{2}} \ln\left| \frac{1 + \sqrt{2}t}{1 - \sqrt{2}t} \right| - \frac{1}{2} \ln\left| \frac{1 + t}{1 - t} \right| \right] + C\)
\(= \frac{1}{4\sqrt{2}} \ln\left| \frac{1 + \sqrt{2}\sin x}{1 - \sqrt{2}\sin x} \right| - \frac{1}{8} \ln\left| \frac{1 + \sin x}{1 - \sin x} \right| + C\)
In simple words: Expand \(\sin 4x\) into multiple half-angles, substitute \(\sin x = t\), and apply partial fractions.

Exam Tip: Partial fraction decomposition is extremely helpful after trigonometric substitutions yield rational functions in terms of sine or cosine.

Question 7. \(\int \frac{dx}{3 + 2 \sin x + \cos x}\)
Answer: Use the Weierstrass substitution (half-angle formulas):
\(\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}, \quad \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}, \quad dx = \frac{2 \, dt}{1 + t^2}\) where \(t = \tan \frac{x}{2}\).
Substitute these into the integral:
\(\int \frac{\frac{2 \, dt}{1 + t^2}}{3 + 2\left(\frac{2t}{1 + t^2}\right) + \frac{1 - t^2}{1 + t^2}} = \int \frac{2 \, dt}{3(1 + t^2) + 4t + (1 - t^2)}\)
\(= \int \frac{2 \, dt}{3 + 3t^2 + 4t + 1 - t^2} = \int \frac{2 \, dt}{2t^2 + 4t + 4} = \int \frac{dt}{t^2 + 2t + 2}\)
Complete the square in the denominator:
\(t^2 + 2t + 2 = (t + 1)^2 + 1\)
The integral becomes:
\(\int \frac{dt}{(t + 1)^2 + 1^2} = \tan^{-1}(t + 1) + C\)
Substitute back \(t = \tan \frac{x}{2}\):
\(= \tan^{-1}\left(\tan \frac{x}{2} + 1\right) + C\)
In simple words: Use half-angle substitutions for sine and cosine to convert the trigonometric integral into a standard rational integral, then complete the square.

Exam Tip: Integrals of the form \(\int \frac{dx}{a + b \sin x + c \cos x}\) are best solved using the substitution \(t = \tan \frac{x}{2}\).

Question 8. \(\int_{0}^{\frac{\pi}{2}} \frac{\sin 2\theta}{\sin^4 \theta + \cos^4 \theta} \, d\theta\)
Answer: Let \(I = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2\theta}{\sin^4 \theta + \cos^4 \theta} \, d\theta\).
Substitute \(\sin^2 \theta = t\), then \(2 \sin \theta \cos \theta \, d\theta = dt\), which means \(\sin 2\theta \, d\theta = dt\).
When \(\theta = 0\), \(t = 0\); when \(\theta = \frac{\pi}{2}\), \(t = 1\).
Also, write \(\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta)^2 + (\cos^2 \theta)^2 = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2t(1-t) = 1 - 2t + 2t^2\).
The integral transforms to:
\(I = \int_{0}^{1} \frac{dt}{1 - 2t + 2t^2} = \int_{0}^{1} \frac{dt}{2(t^2 - t + \frac{1}{2})} = \frac{1}{2} \int_{0}^{1} \frac{dt}{(t - \frac{1}{2})^2 + (\frac{1}{2})^2}\)
Using the standard formula \(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)\):
\(I = \frac{1}{2} \left[ \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{t - \frac{1}{2}}{\frac{1}{2}}\right) \right]_{0}^{1} = \left[ \tan^{-1}(2t - 1) \right]_{0}^{1}\)
Evaluate at the limits:
\(= \tan^{-1}(2(1) - 1) - \tan^{-1}(2(0) - 1) = \tan^{-1}(1) - \tan^{-1}(-1)\)
\(= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\)
In simple words: Substitute \(\sin^2 \theta = t\) to convert the denominator into a quadratic polynomial, then apply standard integration formulas for inverse tangents.

Exam Tip: Look for symmetries and powers of trigonometric functions where substituting \(\sin^2 \theta = t\) simplifies the entire denominator.

Question 9. \(\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx\)
Answer: Let \(I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx\).
Using the definite integral property \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\):
\(I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} - x}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx\)
\(I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} - x}{\cos x + \sin x} \, dx\)
Adding the two expressions for \(I\):
\(2I = \int_{0}^{\frac{\pi}{2}} \frac{x + \frac{\pi}{2} - x}{\sin x + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin x + \cos x} \, dx\)
\(2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sin x + \cos x}\)
To evaluate \(\int \frac{dx}{\sin x + \cos x}\), multiply numerator and denominator by \(\frac{1}{\sqrt{2}}\):
\(\sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \sin\left(x + \frac{\pi}{4}\right)\)
\(\int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)} = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \csc\left(x + \frac{\pi}{4}\right) dx\)
\(= \frac{1}{\sqrt{2}} \left[ \ln\left| \tan\left(\frac{x}{2} + \frac{\pi}{8}\right) \right| \right]_{0}^{\frac{\pi}{2}} = \frac{1}{\sqrt{2}} \left[ \ln\left(\tan \frac{3\pi}{8}\right) - \ln\left(\tan \frac{\pi}{8}\right) \right]\)
Since \(\tan \frac{3\pi}{8} = \sqrt{2} + 1\) and \(\tan \frac{\pi}{8} = \sqrt{2} - 1\):
\(= \frac{1}{\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) = \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1)^2 = \sqrt{2} \ln(\sqrt{2} + 1)\)
Thus, \(2I = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1)\) (or expressed alternatively using standard King's property symmetry reductions).
Standard evaluation yields: \(I = \frac{\pi}{2\sqrt{2}} \ln(\sqrt{2} + 1)\).
In simple words: Apply the King's property of definite integrals (\(\int f(x) = \int f(a+b-x)\)) to eliminate \(x\) from the numerator, then evaluate the resulting standard integral.

Exam Tip: The property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\) is the standard tool for eliminating \(x\) from numerators in trigonometric integrals.

Question 10. \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|) \, dx\)
Answer: Since \(\sin|-x| = -\sin|x|\) and \(\cos|-x| = \cos|x|\), \(\sin|x|\) is an odd function and \(\cos|x|\) is an even function on the symmetric interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Using the property of even and odd functions:
\(\int_{-a}^{a} \text{odd function} \, dx = 0\) and \(\int_{-a}^{a} \text{even function} \, dx = 2 \int_{0}^{a} \text{even function} \, dx\).
Therefore:
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin|x| \, dx = 0\)
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos|x| \, dx = 2 \int_{0}^{\frac{\pi}{2}} \cos x \, dx\) (since for \(x \in [0, \frac{\pi}{2}], |x| = x\))
Evaluate the remaining integral:
\(2 \int_{0}^{\frac{\pi}{2}} \cos x \, dx = 2 [\sin x]_{0}^{\frac{\pi}{2}} = 2 (\sin\frac{\pi}{2} - \sin 0) = 2(1 - 0) = 2\)
Wait, re-evaluating the absolute values directly over the interval: for \(x \in \left[-\frac{\pi}{2}, 0\right]\), \(\sin|x| = \sin(-x) = -\sin x\) and \(\cos|x| = \cos(-x) = \cos x\).
Let us evaluate directly:
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|) \, dx = \int_{-\frac{\pi}{2}}^{0} (-\sin x + \cos x) \, dx + \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx\)
\(= [\cos x + \sin x]_{-\frac{\pi}{2}}^{0} + [-\cos x + \sin x]_{0}^{\frac{\pi}{2}}\)
\(= (\cos 0 + \sin 0) - (\cos(-\frac{\pi}{2}) + \sin(-\frac{\pi}{2})) + (-\cos\frac{\pi}{2} + \sin\frac{\pi}{2}) - (-\cos 0 + \sin 0)\)
\(= (1 + 0) - (0 - 1) + (0 + 1) - (-1 + 0) = 1 + 1 + 1 + 1 = 4\).
In simple words: Split the integral at \(x = 0\) to handle the absolute value definitions, then integrate each part separately.

Exam Tip: Always split integrals containing absolute value functions at points where the expression inside the absolute value changes sign.

Question 11. \(\int \frac{x+1}{x(1+x e^x)^2} \, dx\)
Answer: Rewrite the integrand by multiplying and dividing the numerator and denominator by \(e^x\):
\(\int \frac{x+1}{x(1+x e^x)^2} \, dx = \int \frac{(x+1)e^x}{xe^x(1+xe^x)^2} \, dx\)
Notice that the derivative of \(xe^x\) is \((x+1)e^x\). Let \(t = xe^x\), then \(dt = (x+1)e^x \, dx\).
Substitute into the integral:
\(\int \frac{dt}{t(1+t)^2}\)
Using partial fractions for \(\frac{1}{t(1+t)^2}\):
\(\frac{1}{t(1+t)^2} = \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2}\)
Integrating each term with respect to \(t\):
\(\int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt = \ln|t| - \ln|1+t| + \frac{1}{1+t} + C\)
Using logarithm properties: \(\ln\left| \frac{t}{1+t} \right| + \frac{1}{1+t} + C\)
Substitute back \(t = xe^x\):
\(= \ln\left| \frac{xe^x}{1+xe^x} \right| + \frac{1}{1+xe^x} + C\)
In simple words: Multiply numerator and denominator by \(e^x\) to create the derivative of \(xe^x\) in the numerator, then use partial fractions.

Exam Tip: Introducing \(e^x\) into rational expressions involving \(x\) and exponentials is a standard trick to enable successful substitutions.

Question 12. \(\int \sqrt{\cot x} \, dx\)
Answer: Let \(t^2 = \cot x\), then \(2t \, dt = -\csc^2 x \, dx = -(1 + \cot^2 x) \, dx = -(1 + t^4) \, dx\), which means \(dx = \frac{-2t}{1 + t^4} \, dt\).
The integral becomes:
\(\int t \left( \frac{-2t}{1 + t^4} \right) dt = -2 \int \frac{t^2}{1 + t^4} \, dt\)
Divide numerator and denominator by \(t^2\):
\(-2 \int \frac{1}{t^2 + \frac{1}{t^2}} \, dt = -2 \int \frac{1}{(t - \frac{1}{t})^2 + 2} \, dt\)
Let \(u = t - \frac{1}{t}\), then \(du = \left(1 + \frac{1}{t^2}\right) dt\). Also, note that \(t^2 + \frac{1}{t^2} = u^2 + 2\), but we can split \(\frac{t^2}{1+t^4}\) as:
\(\frac{t^2}{1+t^4} = \frac{1}{2} \left[ \frac{t^2 + 1}{1 + t^4} + \frac{t^2 - 1}{1 + t^4} \right]\)
Integrating the two parts separately using substitutions \(t - \frac{1}{t} = u\) and \(t + \frac{1}{t} = v\):
Standard result for \(\int \sqrt{\cot x} \, dx\):
\(= -\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\cot x - 1}{\sqrt{2 \cot x}}\right) - \frac{1}{2\sqrt{2}} \ln\left| \frac{\cot x - \sqrt{2\cot x} + 1}{\cot x + \sqrt{2\cot x} + 1} \right| + C\)
In simple words: Substitute \(t^2 = \cot x\), then use algebraic manipulations of numerator and denominator to split into standard inverse tangent and logarithmic integrals.

Exam Tip: Integrals of square roots of trigonometric functions like \(\sqrt{\tan x}\) or \(\sqrt{\cot x}\) always require substituting \(t^2 = \text{trig function}\) followed by splitting into sum and difference terms.