CBSE Class 12 Mathematics Integrals VBQs Set B

Fill in the Blanks

Question. \( \int_0^{\pi/2} \frac{\sin^n x \, dx}{\sin^n x + \cos^n x} = \) _____________ .
Answer: \( \frac{\pi}{4} \)

Question. \( \int_0^{\pi/2} \cos x e^{\sin x} dx = \) _____________ .
Answer: \( e - 1 \)

Question. \( \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx = \) _____________ .
Answer: \( x e^{\tan^{-1} x} + C \)

Question. A primitive of \( |x| \), when \( x < 0 \) is _____________ .
Answer: \( -\frac{1}{2} x^2 + C \)

Question. The value of \( \int_{-\pi}^\pi \sin^3 x \cos^2 x \, dx = \) _____________ .
Answer: 0

Very Short Answer Questions

Question. Evaluate: \( \int \frac{dx}{9 + 4x^2} \)
Answer: \( \int \frac{dx}{9 + 4x^2} = \frac{1}{4} \int \frac{dx}{\frac{9}{4} + x^2} = \frac{1}{4} \int \frac{dx}{(\frac{3}{2})^2 + x^2} \)
\( = \frac{1}{4} \cdot \frac{1}{3/2} \tan^{-1} (\frac{x}{3/2}) + C \)
\( = \frac{1}{6} \tan^{-1} (\frac{2x}{3}) + C \)

Question. Find: \( \int \frac{2^{x+1} - 5^{x-1}}{10^x} dx \)
Answer: \( \int \frac{2^{x+1} - 5^{x-1}}{10^x} dx = \int \frac{2^{x+1}}{(5 \times 2)^x} dx - \int \frac{5^{x-1}}{(5 \times 2)^x} dx \)
\( = \int \frac{2^x \times 2}{5^x \times 2^x} dx - \int \frac{5^x \times \frac{1}{5}}{5^x \times 2^x} dx \)
\( = 2 \int 5^{-x} dx - \frac{1}{5} \int 2^{-x} dx = \frac{-2 \times 5^{-x}}{\log 5} - \frac{1}{5} \times \frac{-2^{-x}}{\log 2} + C \)
\( = \frac{-2}{\log 5} 5^{-x} + \frac{1}{5} \times \frac{2^{-x}}{\log 2} + C \)
\( = \frac{1}{5 \log 2 \cdot 2^x} - \frac{2}{\log 5 \cdot 5^x} + C \)

Question. \( \int \frac{(x^2 + 2)}{x + 1} dx \)
Answer: Let \( I = \int \frac{x^2 + 2}{x + 1} dx = \int (x - 1 + \frac{3}{x + 1}) dx \)
\( = \int (x - 1) dx + 3 \int \frac{1}{x + 1} dx \)
\( = \frac{x^2}{2} - x + 3 \log |x + 1| + C \)

Question. Evaluate: \( \int \sec^2 (7 - x) dx \)
Answer: \( \int \sec^2 (7 - x) dx = \frac{\tan (7 - x)}{-1} + C = -\tan (7 - x) + C \)

Question. Evaluate: \( \int \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx \)
Answer: Let \( \sqrt{x} = z \Rightarrow \frac{1}{2\sqrt{x}} dx = dz \Rightarrow \frac{dx}{\sqrt{x}} = 2dz \)
\( \therefore \int \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx = 2 \int \sec^2 z \, dz = 2 \tan z + C = 2 \tan \sqrt{x} + C \)

Question. Evaluate: \( \int \frac{dx}{x + x \log x} \)
Answer: Let \( I = \int \frac{dx}{x(1 + \log x)} \)
Put \( 1 + \log x = z \Rightarrow \frac{1}{x} dx = dz \)
\( \therefore I = \int \frac{dz}{z} = \log |z| + C = \log |1 + \log x| + C \)

Question. If \( \int_0^a \frac{dx}{1 + 4x^2} = \frac{\pi}{8} \), then find the value of \( a \).
Answer: We have, \( \int_0^a \frac{dx}{1 + 4x^2} = \frac{\pi}{8} \)
\( \Rightarrow \frac{1}{4} \int_0^a \frac{dx}{\frac{1}{4} + x^2} = \frac{\pi}{8} \Rightarrow \frac{1}{4} \int_0^a \frac{dx}{(1/2)^2 + x^2} = \frac{\pi}{8} \)
\( \Rightarrow \frac{1}{4} \times \left[ \frac{1}{1/2} \tan^{-1} \frac{x}{1/2} \right]_0^a = \frac{\pi}{8} \)
\( \Rightarrow \frac{1}{4} \times 2 [ \tan^{-1} 2a - \tan^{-1} 0 ] = \frac{\pi}{8} \)
\( \Rightarrow \tan^{-1} 2a = \frac{\pi}{4} \Rightarrow 2a = \tan \frac{\pi}{4} = 1 \Rightarrow a = \frac{1}{2} \)

Question. Find the value of \( \int_1^4 |x - 5| dx \)
Answer: We have, \( \int_1^4 |x - 5| dx = \int_1^4 -(x - 5) dx \)
\( = - [\frac{x^2}{2} - 5x]_1^4 = - [(\frac{4^2}{2} - 5 \times 4) - (\frac{1^2}{2} - 5 \times 1)] \)
\( = - [ (8 - 20) - (1/2 - 5) ] = - [ -12 + 4.5 ] = 7.5 = \frac{15}{2} \)

Question. If \( \int_0^1 (3x^2 + 2x + k) dx = 0 \), then find the value of \( k \).
Answer: Given, \( \int_0^1 (3x^2 + 2x + k) dx = 0 \Rightarrow [x^3 + x^2 + kx]_0^1 = 0 \)
\( \Rightarrow (1 + 1 + k) - (0) = 0 \Rightarrow k = -2 \)

Short Answer Questions-I 

Question. Find \( \int \frac{\tan^3 x}{\cos^3 x} dx \)
Answer: We have, \( I = \int \frac{\tan^3 x}{\cos^3 x} dx = \int \frac{\sin^3 x}{\cos^3 x} \cdot \frac{1}{\cos^3 x} dx = \int \frac{\sin^3 x}{\cos^6 x} dx \)
Let \( \cos x = t \Rightarrow - \sin x \, dx = dt \Rightarrow \sin x \, dx = -dt \)
\( \therefore I = \int \frac{\sin^2 x \sin x \, dx}{\cos^6 x} = \int \frac{(1 - \cos^2 x) \sin x \, dx}{\cos^6 x} \)
\( = \int \frac{-(1 - t^2) dt}{t^6} = \int \frac{t^2 - 1}{t^6} dt = \int (t^{-4} - t^{-6}) dt \)
\( = \frac{t^{-3}}{-3} - \frac{t^{-5}}{-5} + C = \frac{1}{5t^5} - \frac{1}{3t^3} + C \)
\( = \frac{1}{5 (\cos x)^5} - \frac{1}{3 (\cos x)^3} + C = \frac{1}{5} \sec^5 x - \frac{1}{3} \sec^3 x + C \)

Question. Find \( \int e^x \frac{\sqrt{1 + \sin 2x}}{1 + \cos 2x} dx \).
Answer: \( I = \int e^x \frac{\sqrt{1 + \sin 2x}}{1 + \cos 2x} dx = \int e^x \frac{\sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x}}{2 \cos^2 x} dx \)
\( = \int e^x \frac{\sqrt{(\sin x + \cos x)^2}}{2 \cos^2 x} dx = \int e^x \frac{\sin x + \cos x}{2 \cos^2 x} dx \)
\( = \frac{1}{2} \int e^x (\frac{\sin x}{\cos^2 x} + \frac{\cos x}{\cos^2 x}) dx = \frac{1}{2} \int e^x (\sec x \tan x + \sec x) dx \)
\( = \frac{1}{2} e^x \sec x + C \) \( [\because \int e^x (f(x) + f'(x)) dx = e^x f(x) + C] \)

Question. Find \( \int \frac{x - 1}{(x - 2)(x - 3)} dx \).
Answer: \( \frac{x - 1}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \)
where \( A = \left. \frac{x-1}{x-3} \right|_{x=2} = \frac{1}{-1} = -1 \) & \( B = \left. \frac{x-1}{x-2} \right|_{x=3} = \frac{2}{1} = 2 \)
\( \therefore \frac{x - 1}{(x - 2)(x - 3)} = \frac{-1}{x - 2} + \frac{2}{x - 3} \)
\( \Rightarrow \int \frac{x - 1}{(x - 2)(x - 3)} dx = - \int \frac{dx}{x - 2} + 2 \int \frac{dx}{x - 3} = - \log (x - 2) + 2 \log (x - 3) + C \)
\( = \log (x - 3)^2 - \log (x - 2) + C = \log \frac{(x - 3)^2}{(x - 2)} + C \)

Question. Find \( \int_{-\pi/4}^0 \frac{1 + \tan x}{1 - \tan x} dx \).
Answer: \( \int_{-\pi/4}^0 \frac{1 + \tan x}{1 - \tan x} dx = \int_{-\pi/4}^0 \tan(\frac{\pi}{4} + x) dx \)
\( = [\log \sec (\frac{\pi}{4} + x)]_{-\pi/4}^0 = \log \sec (\frac{\pi}{4}) - \log \sec (\frac{\pi}{4} - \frac{\pi}{4}) \)
\( = \log (\sqrt{2}) - \log (\sec 0) = \log \sqrt{2} - \log 1 \)
\( = \log \sqrt{2} = \frac{1}{2} \log 2 \)

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

Question. Evaluate \( \int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx \).
Answer: Let \( t = \tan^{-1} x \Rightarrow dt = \frac{1}{1 + x^2} dx \)
Also when, \( x = 0, t = 0 \) and when \( x = 1, t = \frac{\pi}{4} \)
\( \therefore \int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx = \int_0^{\pi/4} t dt = \left[ \frac{t^2}{2} \right]_0^{\pi/4} = \frac{1}{2} [ \frac{\pi^2}{16} - 0 ] = \frac{\pi^2}{32} \)

Question. Evaluate: \( \int_0^1 \frac{dx}{\sqrt{2x + 3}} \).
Answer: Let \( I = \int_0^1 (2x + 3)^{-1/2} dx \)
\( = \left[ \frac{(2x + 3)^{-1/2+1}}{(-1/2 + 1) \times 2} \right]_0^1 = \left[ \frac{(2x + 3)^{1/2}}{1} \right]_0^1 = \sqrt{5} - \sqrt{3} \)

Short Answer Questions-II 

Question. Evaluate: \( \int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx \)
Answer: Let \( x^2 = z \Rightarrow 2x dx = dz \)
\( \therefore \int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx = \int \frac{dz}{(z + 1)(z + 3)} \)
Using partial fraction.
Let \( \frac{1}{(z + 1)(z + 3)} = \frac{A}{z + 1} + \frac{B}{z + 3} \) ...(i)
\( \frac{1}{(z + 1)(z + 3)} = \frac{A(z + 3) + B(z + 1)}{(z + 1)(z + 3)} \)
\( \Rightarrow 1 = A(z + 3) + B(z + 1) \Rightarrow 1 = (A + B)z + (3A + B) \)
Equating the coefficient of \( z \) and constant, we get
\( A + B = 0 \) ...(ii)
and \( 3A + B = 1 \) ...(iii)
Subtracting (ii) from (iii), we get \( 2A = 1 \Rightarrow A = 1/2 \) and \( B = -1/2 \)
Putting the values of \( A \) and \( B \) in (i), we get
\( \frac{1}{(z + 1)(z + 3)} = \frac{1}{2(z + 1)} - \frac{1}{2(z + 3)} \)
\( \therefore \int \frac{2x dx}{(x^2 + 1)(x^2 + 3)} = \int \left( \frac{1}{2(z + 1)} - \frac{1}{2(z + 3)} \right) dz = \frac{1}{2} \int \frac{dz}{z + 1} - \frac{1}{2} \int \frac{dz}{z + 3} \)
\( = \frac{1}{2} \log |z + 1| - \frac{1}{2} \log |z + 3| + C = \frac{1}{2} \log |x^2 + 1| - \frac{1}{2} \log |x^2 + 3| + C \)
\( = \frac{1}{2} \log \frac{x^2 + 1}{x^2 + 3} + C = \log \sqrt{\frac{x^2 + 1}{x^2 + 3}} + C \)

Question. Evaluate: \( \int \frac{x^2}{1 - x^4} dx \)
Answer: Let \( I = \int \frac{x^2}{1 - x^4} dx = \int \frac{\frac{1}{2} (x^2 + 1) - \frac{1}{2} (1 - x^2)}{(1 - x^2)(1 + x^2)} dx \)
\( = \int \frac{\frac{1}{2}(1 + x^2)}{(1 - x^2)(1 + x^2)} dx - \frac{1}{2} \int \frac{(1 - x^2)}{(1 - x^2)(1 + x^2)} dx \)
\( = \frac{1}{2} \int \frac{1}{1 - x^2} dx - \frac{1}{2} \int \frac{1}{1 + x^2} dx = \frac{1}{2} \cdot \frac{1}{2} \log \left| \frac{1 + x}{1 - x} \right| + C_1 - \frac{1}{2} \tan^{-1} x + C_2 \)
\( = \frac{1}{4} \log \left| \frac{1 + x}{1 - x} \right| - \frac{1}{2} \tan^{-1} x + C \)

Question. Evaluate: \( \int \sin x \sin 2x \sin 3x \, dx \)
Answer: Let \( I = \int \sin x \sin 2x \sin 3x \, dx \)
\( = \frac{1}{2} \int \sin x \cdot (2 \sin 2x \sin 3x) dx = \frac{1}{2} \int \sin x \cdot (\cos x - \cos 5x) dx \)
\( = \frac{1}{2 \times 2} \int 2 \sin x \cos x \, dx - \frac{1}{2 \times 2} \int 2 \sin x \cos 5x \, dx \)
\( = \frac{1}{4} \int \sin 2x \, dx - \frac{1}{4} \int (\sin 6x - \sin 4x) dx \)
\( = - \frac{\cos 2x}{8} + \frac{\cos 6x}{24} + \frac{\cos 4x}{16} + C \)

Question. Evaluate: \( \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \)
Answer: Let \( I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx \Rightarrow I = \int \frac{(\sin^2 x)^3 + (\cos^2 x)^3}{\sin^2 x \cos^2 x} dx \)
\( \Rightarrow I = \int \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)}{\sin^2 x \cos^2 x} dx \)
\( \Rightarrow I = \int \frac{\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x}{\sin^2 x \cos^2 x} dx = \int \tan^2 x \, dx - \int dx + \int \cot^2 x \, dx \)
\( \Rightarrow I = \int (\sec^2 x - 1) dx - \int dx + \int (\csc^2 x - 1) dx \)
\( \Rightarrow I = \int \sec^2 x \, dx + \int \csc^2 x \, dx - x - x - x + C = \tan x - \cot x - 3x + C \)

Question. Evaluate: \( \int \frac{\sin(x - a)}{\sin(x + a)} dx \)
Answer: Let \( I = \int \frac{\sin(x - a)}{\sin(x + a)} dx \)
Let \( x + a = t \Rightarrow x = t - a \Rightarrow dx = dt \)
\( \therefore I = \int \frac{\sin(t - 2a)}{\sin t} dt = \int \frac{\sin t \cos 2a - \cos t \sin 2a}{\sin t} dt \)
\( = \cos 2a \int dt - \int \sin 2a \cot t \, dt = \cos 2a \cdot t - \sin 2a \cdot \log |\sin t| + C \)
\( = \cos 2a \cdot (x + a) - \sin 2a \cdot \log |\sin (x + a)| + C \)
\( = x \cos 2a + a \cos 2a - (\sin 2a) \log |\sin (x + a)| + C \)

Question. Evaluate: \( \int \frac{e^x}{\sqrt{5 - 4e^x - e^{2x}}} dx \)
Answer: Let \( I = \int \frac{e^x}{\sqrt{5 - 4e^x - e^{2x}}} dx \)
Put \( e^x = t \Rightarrow e^x dx = dt \), we get
\( \therefore I = \int \frac{dt}{\sqrt{5 - 4t - t^2}} = \int \frac{dt}{\sqrt{-(t^2 + 4t - 5)}} = \int \frac{dt}{\sqrt{-(t^2 + 2.t.2 + 2^2 - 9)}} \)
\( = \int \frac{dt}{\sqrt{3^2 - (t + 2)^2}} = \sin^{-1} \frac{t + 2}{3} + C = \sin^{-1} (\frac{e^x + 2}{3}) + C \)

Question. Evaluate: \( \int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx \)
Answer: Let \( I = \int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx \)
\( = \int e^x \left( \frac{2 \sin 2x \cos 2x - 4}{2 \sin^2 2x} \right) dx \)
\( = \int e^x (\cot 2x - 2 \csc^2 2x) dx \)
Let \( f(x) = \cot 2x \therefore f'(x) = -2 \csc^2 2x \)
\( \therefore I = \int e^x (f(x) + f'(x)) dx = e^x \cdot f(x) + C = e^x \cdot \cot 2x + C \)

Question. Evaluate: \( \int \frac{x + 2}{\sqrt{x^2 + 5x + 6}} dx \)
Answer: Let \( I = \int \frac{x + 2}{\sqrt{x^2 + 5x + 6}} dx \)
Now, we can express as \( x + 2 = A \frac{d}{dx}(x^2 + 5x + 6) + B \)
\( \Rightarrow x + 2 = A(2x + 5) + B \Rightarrow x + 2 = 2Ax + (5A + B) \)
Equating coefficients both sides, we get
\( 2A = 1, 5A + B = 2 \Rightarrow A = 1/2, B = 2 - 5/2 = -1/2 \)
\( \therefore x + 2 = \frac{1}{2}(2x + 5) - \frac{1}{2} \)
Hence, \( I = \int \frac{\frac{1}{2}(2x + 5) - \frac{1}{2}}{\sqrt{x^2 + 5x + 6}} dx = \frac{1}{2} \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx - \frac{1}{2} \int \frac{dx}{\sqrt{x^2 + 5x + 6}} \)
\( I = \frac{1}{2} I_1 - \frac{1}{2} I_2 \) ... (i)
where, \( I_1 = \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx, I_2 = \int \frac{dx}{\sqrt{x^2 + 5x + 6}} \)
Now, \( I_1 = \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx \)
Let \( x^2 + 5x + 6 = z \Rightarrow (2x + 5) dx = dz \)
\( \therefore I_1 = \int \frac{dz}{\sqrt{z}} = \int z^{-1/2} dz = \frac{z^{1/2}}{1/2} + C_1 = 2\sqrt{z} + C_1 = 2\sqrt{x^2 + 5x + 6} + C_1 \)
Again \( I_2 = \int \frac{dx}{\sqrt{x^2 + 2 \times x \times \frac{5}{2} + (\frac{5}{2})^2 - \frac{25}{4} + 6}} = \int \frac{dx}{\sqrt{(x + \frac{5}{2})^2 - \frac{1}{4}}} = \int \frac{dx}{\sqrt{(x + \frac{5}{2})^2 - (\frac{1}{2})^2}} \)
\( = \log |(x + \frac{5}{2}) + \sqrt{x^2 + 5x + 6}| + C_2 \)
Putting the value of \( I_1 \) and \( I_2 \) in (i), we get
\( I = \frac{1}{2} [2\sqrt{x^2 + 5x + 6} + C_1] - \frac{1}{2} \{ \log |(x + \frac{5}{2}) + \sqrt{x^2 + 5x + 6}| + C_2 \} \)
\( = \sqrt{x^2 + 5x + 6} - \frac{1}{2} \log |(x + \frac{5}{2}) + \sqrt{x^2 + 5x + 6}| + \frac{1}{2} C_1 - \frac{1}{2} C_2 \)
\( = \sqrt{x^2 + 5x + 6} - \frac{1}{2} \log |(x + \frac{5}{2}) + \sqrt{x^2 + 5x + 6}| + C \)

Question. Evaluate: \( \int \frac{(x^2 - 3x)}{(x - 1)(x - 2)} dx \)
Answer: Let \( I = \int \frac{x^2 - 3x}{(x - 1)(x - 2)} dx = \int \frac{x^2 - 3x}{x^2 - 3x + 2} dx \)
\( = \int \frac{x^2 - 3x + 2 - 2}{x^2 - 3x + 2} dx = \int dx - \int \frac{2 dx}{x^2 - 3x + 2} \)
\( = x - 2 \int \frac{dx}{x^2 - 2 \cdot x \cdot \frac{3}{2} + \frac{9}{4} - \frac{9}{4} + 2} = x - 2 \int \frac{dx}{(x - \frac{3}{2})^2 - (\frac{1}{2})^2} \)
\( = x - 2 \left[ \frac{1}{2(1/2)} \log \left| \frac{x - \frac{3}{2} - \frac{1}{2}}{x - \frac{3}{2} + \frac{1}{2}} \right| \right] + C \)
\( = x - 2 \log \left| \frac{x - 2}{x - 1} \right| + C \)

Question. Find: \( \int \sin^{-1} \sqrt{\frac{x}{a + x}} dx \)
Answer: Let \( I = \int \sin^{-1} \sqrt{\frac{x}{a + x}} dx \)
Put \( x = a \tan^2 \theta \Rightarrow dx = 2a \tan \theta \sec^2 \theta \, d\theta \)
\( \therefore I = \int \sin^{-1} \left( \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} \right) (2a \tan \theta \sec^2 \theta) d\theta \)
\( = \int \sin^{-1} (\frac{\tan \theta}{\sec \theta}) 2a \tan \theta \sec^2 \theta \, d\theta = \int \sin^{-1} (\sin \theta) 2a \tan \theta \sec^2 \theta \, d\theta \)
\( = 2a \int \theta \cdot \tan \theta \sec^2 \theta \, d\theta \)
\( = 2a \left[ \theta \cdot \int \tan \theta \sec^2 \theta \, d\theta - \int \left( \frac{d}{d\theta} \theta \cdot \int \tan \theta \sec^2 \theta \, d\theta \right) d\theta \right] \)
\( = 2a \left[ \theta \cdot \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} d\theta \right] = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta \)
\( = a\theta \tan^2 \theta - a \tan \theta + a\theta + C \)
\( = a \frac{x}{a} \tan^{-1} \sqrt{\frac{x}{a}} - a \sqrt{\frac{x}{a}} + a \tan^{-1} \sqrt{\frac{x}{a}} + C \)
\( = (x + a) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + C \)

Question. Find : \( \int \frac{dx}{\sin x + \sin 2x} \)
Answer: Here, \( I = \int \frac{1}{\sin x + 2 \sin x \cos x} dx \)
\( \Rightarrow I = \int \frac{1}{\sin x (1 + 2 \cos x)} dx \Rightarrow I = \int \frac{\sin x}{\sin^2 x (1 + 2 \cos x)} dx \)
\( \Rightarrow I = \int \frac{\sin x}{(1 - \cos^2 x)(1 + 2 \cos x)} dx = \int \frac{\sin x}{(1 - \cos x)(1 + \cos x)(1 + 2 \cos x)} dx \)
Let \( \cos x = z \Rightarrow - \sin x \, dx = dz \)
\( \Rightarrow I = - \int \frac{dz}{(1 - z)(1 + z)(1 + 2z)} \)
Integrand is proper rational function. By form of partial fraction:
\( \frac{1}{(1 - z)(1 + z)(1 + 2z)} = \frac{A}{1 - z} + \frac{B}{1 + z} + \frac{C}{1 + 2z} \) ...(i)
\( \Rightarrow 1 = A(1 + z)(1 + 2z) + B(1 - z)(1 + 2z) + C(1 + z)(1 - z) \) ...(ii)
Putting \( z = 1 \Rightarrow 1 = A(2)(3) \Rightarrow A = 1/6 \)
Putting \( z = -1 \Rightarrow 1 = B(2)(-1) \Rightarrow B = -1/2 \)
Putting \( z = -1/2 \Rightarrow 1 = C(1/2)(3/2) \Rightarrow C = 4/3 \)
Putting values in (i), we get \( \frac{1}{(1 - z)(1 + z)(1 + 2z)} = \frac{1/6}{1 - z} + \frac{-1/2}{1 + z} + \frac{4/3}{1 + 2z} \)
\( \therefore I = - \int [ \frac{1}{6(1 - z)} - \frac{1}{2(1 + z)} + \frac{4}{3(1 + 2z)} ] dz = \frac{1}{6} \int \frac{dz}{z - 1} + \frac{1}{2} \int \frac{dz}{1 + z} - \frac{4}{3} \int \frac{dz}{1 + 2z} \)
\( \Rightarrow I = \frac{1}{6} \log |z - 1| + \frac{1}{2} \log |1 + z| - \frac{4}{3 \times 2} \log |1 + 2z| + C \)
Putting \( z = \cos x \):
\( \Rightarrow I = \frac{1}{6} \log |1 - \cos x| + \frac{1}{2} \log |1 + \cos x| - \frac{2}{3} \log |1 + 2 \cos x| + C \)

Question. Find: \( \int \frac{x^2}{x^4 - x^2 - 12} dx \)
Answer: Let \( I = \int \frac{x^2}{x^4 - x^2 - 12} dx = \int \frac{x^2}{(x^2 - 4)(x^2 + 3)} dx \)
Let \( x^2 = t \). Then \( \frac{t}{(t - 4)(t + 3)} = \frac{A}{t - 4} + \frac{B}{t + 3} \)
\( t = A(t + 3) + B(t - 4) \). Comparing coeff. of \( t \): \( A + B = 1 \). Constant: \( 3A - 4B = 0 \).
\( \Rightarrow B = 3/4 A \Rightarrow A + 3/4 A = 1 \Rightarrow 7/4 A = 1 \Rightarrow A = 4/7, B = 3/7 \)
\( \therefore \frac{x^2}{(x^2 - 4)(x^2 + 3)} = \frac{4}{7(x^2 - 4)} + \frac{3}{7(x^2 + 3)} \)
\( I = \frac{4}{7} \int \frac{dx}{x^2 - 2^2} + \frac{3}{7} \int \frac{dx}{x^2 + (\sqrt{3})^2} \)
\( = \frac{4}{7} \cdot \frac{1}{2 \times 2} \log \left| \frac{x - 2}{x + 2} \right| + \frac{3}{7} \cdot \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} + C \)
\( = \frac{1}{7} \log \left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{-1} \frac{x}{\sqrt{3}} + C \)

Question. Evaluate : \( \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} dx \)
Answer: Let \( I = \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} dx \). Put \( x^2 = t \).
\( \frac{t}{(t + 4)(t + 9)} = \frac{A}{t + 4} + \frac{B}{t + 9} \Rightarrow t = A(t + 9) + B(t + 4) \)
Equating coefficients: \( A + B = 1, 9A + 4B = 0 \).
Solving: \( A = -4/5, B = 9/5 \).
\( \therefore \frac{x^2}{(x^2 + 4)(x^2 + 9)} = -\frac{4}{5(x^2 + 4)} + \frac{9}{5(x^2 + 9)} \)
\( \Rightarrow \int \frac{x^2 dx}{(x^2 + 4)(x^2 + 9)} = -\frac{4}{5} \int \frac{dx}{x^2 + 2^2} + \frac{9}{5} \int \frac{dx}{x^2 + 3^2} = -\frac{4}{5} \times \frac{1}{2} \tan^{-1} \frac{x}{2} + \frac{9}{5} \times \frac{1}{3} \tan^{-1} \frac{x}{3} + C \)
\( = -\frac{2}{5} \tan^{-1} \frac{x}{2} + \frac{3}{5} \tan^{-1} \frac{x}{3} + C \)