CBSE Class 12 Mathematics Integrals Worksheet Set 12

Long Answer Type Questions

Question. Evaluate: \( \int \sin x \sin 2x \sin 3x \, dx \)
Answer: Let \( I = \int \sin x \sin 2x \sin 3x \, dx \)
\( = \frac{1}{2} \int 2 \sin x \sin 2x \sin 3x \, dx = \frac{1}{2} \int \sin x \cdot (2 \sin 2x \sin 3x) \, dx \)
\( = \frac{1}{2} \int \sin x \cdot (\cos x - \cos 5x) \, dx \quad [\because 2 \sin A \sin B = \cos(A - B) - \cos(A + B)] \)
\( = \frac{1}{2 \times 2} \int 2 \sin x \cos x \, dx - \frac{1}{2 \times 2} \int 2 \sin x \cos 5x \, dx \)
\( = \frac{1}{4} \int \sin 2x \, dx - \frac{1}{4} \int (\sin 6x - \sin 4x) \, dx \quad \left[ \because \sin C - \sin D = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2} \right] \)
\( = -\frac{\cos 2x}{8} + \frac{\cos 6x}{24} - \frac{\cos 4x}{16} + C \)

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

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

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 \( e^x = t \)
\( \implies \) \( e^x \, dx = dt \), we get
\( \therefore I = \int \frac{dt}{\sqrt{5 - 4t - t^2}} = \int \frac{dt}{\sqrt{-(t^2 + 4t - 5)}} = \int \frac{dt}{\sqrt{-(t^2 + 2.t.2 + 2^2 - 2^2 - 5)}} \)
\( = \int \frac{dt}{\sqrt{3^2 - (t + 2)^2}} = \sin^{-1} \frac{t + 2}{3} + C = \sin^{-1} \left( \frac{e^x + 2}{3} \right) + C \)

Question. Evaluate: \( \int x \sin^{-1} x \, dx \)
Answer: Let \( I = \int x \sin^{-1} x \, dx \)
\( = \sin^{-1} x \cdot \frac{x^2}{2} - \int \frac{x^2}{2\sqrt{1 - x^2}} \, dx \quad [\text{By using integration by parts}] \)
\( = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \int \frac{1 - x^2 - 1}{\sqrt{1 - x^2}} \, dx = \frac{x^2}{2} \sin^{-1} x + \frac{1}{2} \int \sqrt{1 - x^2} \, dx - \frac{1}{2} \int \frac{dx}{\sqrt{1 - x^2}} \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \sin^{-1} x + \frac{1}{2} \left[ \frac{x}{2} \sqrt{1 - x^2} + \frac{1}{2} \sin^{-1} x \right] + C \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \sin^{-1} x + \frac{x}{4} \sqrt{1 - x^2} + \frac{1}{4} \sin^{-1} x + C \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{x}{4} \sqrt{1 - x^2} + C \)

Question. Evaluate: \( \int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) \, dx \)
Answer: Let \( I = \int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) \, dx \)
\( = \int e^x \left( \frac{2 \sin 2x \cos 2x - 4}{2 \sin^2 2x} \right) \, dx \quad [\because \sin 2x = 2 \sin x \cos x \text{ and } \cos 2x = 1 - 2 \sin^2 x] \)
\( = \int e^x (\cot 2x - 2 \csc^2 2x) \, dx \)
Let \( f(x) = \cot 2x \)
\( \therefore f'(x) = -2 \csc^2 2x \)
\( \therefore I = \int e^x (f(x) + f'(x)) \, dx \)

\( \implies \) \( I = e^x \cdot f(x) + C = e^x \cdot \cot 2x + C \quad [\because \int e^x(f(x) + f'(x)) \, dx = e^x f(x) + C] \)

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

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

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

Question. Find: \( \int \frac{dx}{\sin x + \sin 2x} \)
Answer: Here, \( I = \int \frac{1}{\sin x + \sin 2x} \, dx \)
\( \implies \) \( I = \int \frac{1}{\sin x + 2 \sin x \cos x} \, dx \)
\( \implies \) \( I = \int \frac{1}{\sin x (1 + 2 \cos x)} \, dx \)
\( \implies \) \( I = \int \frac{\sin x}{\sin^2 x (1 + 2 \cos x)} \, dx \)
\( \implies \) \( I = \int \frac{\sin x}{(1 - \cos^2 x)(1 + 2 \cos x)} \, dx \)
Let \( \cos x = z \)
\( \implies \) \( -\sin x \, dx = dz \)
\( \implies \) \( I = \int \frac{-dz}{(1 - z^2)(1 + 2z)} \)
\( \implies \) \( I = -\int \frac{dz}{(1 + z)(1 - z)(1 + 2z)} \)
Here, integrand is proper rational function. Therefore, by the form of partial function, we can write
\( \frac{1}{(1 + z)(1 - z)(1 + 2z)} = \frac{A}{1 + z} + \frac{B}{1 - z} + \frac{C}{1 + 2z} \quad ...(i) \)
\( \implies \) \( \frac{1}{(1 + z)(1 - z)(1 + 2z)} = \frac{A(1 - z)(1 + 2z) + B(1 + z)(1 + 2z) + C(1 + z)(1 - z)}{(1 + z)(1 - z)(1 + 2z)} \)
\( \implies \) \( 1 = A(1 - z)(1 + 2z) + B(1 + z)(1 + 2z) + C(1 + z)(1 - z) \quad ...(ii) \)
Putting the value of \( z = -1 \) in (ii), we get
\( \implies \) \( 1 = -2A + 0 + 0 \)
\( \implies \) \( A = -1/2 \)
Again, putting the value of \( z = 1 \) in (ii), we get
\( \implies \) \( 1 = 0 + B \cdot 2 \cdot (1 + 2) + 0 \)
\( \implies \) \( 1 = 6B \)
\( \implies \) \( B = \frac{1}{6} \)
Similarly, putting the value of \( z = -\frac{1}{2} \) in (ii), we get
\( \implies \) \( 1 = 0 + 0 + C \left( \frac{1}{2} \right) \left( \frac{3}{2} \right) \)
\( \implies \) \( 1 = \frac{3}{4} C \)
\( \implies \) \( C = \frac{4}{3} \)
Putting the value of \( A, B, C \) in (i), we get
\( \frac{1}{(1 + z)(1 - z)(1 + 2z)} = \frac{-1}{2(1 + z)} + \frac{1}{6(1 - z)} + \frac{4}{3(1 + 2z)} \)
\( \therefore I = - \int \left[ \frac{-1}{2(1 + z)} + \frac{1}{6(1 - z)} + \frac{4}{3(1 + 2z)} \right] \, dz = \int \left[ \frac{1}{2(1 + z)} - \frac{1}{6(1 - z)} - \frac{4}{3(1 + 2z)} \right] \, dz \)
\( \implies \) \( I = \frac{1}{2} \log |1 + z| + \frac{1}{6} \log |1 - z| - \frac{4}{3 \times 2} \log |1 + 2z| + C \)
Putting the value of z, we get
\( \implies \) \( I = \frac{1}{2} \log |1 + \cos x| + \frac{1}{6} \log |1 - \cos x| - \frac{2}{3} \log |1 + 2 \cos x| + C \)

Question. Evaluate: \( \int \frac{dx}{x(x^5 + 3)} \)
Answer: Let \( I = \int \frac{dx}{x(x^5 + 3)} = \int \frac{x^4 \, dx}{x^5(x^5 + 3)} = \frac{1}{5} \int \frac{5x^4 \, dx}{x^5(x^5 + 3)} \)
Put \( x^5 = z \)
\( \implies \) \( 5x^4 \, dx = dz \)
\( \therefore I = \frac{1}{5} \int \frac{dz}{z(z + 3)} = \frac{1}{5 \times 3} \int \frac{z + 3 - z}{z(z + 3)} \, dz = \frac{1}{15} \int \left( \frac{1}{z} - \frac{1}{z + 3} \right) \, dz = \frac{1}{15} [\log z - \log |z + 3|] + C \)
\( \implies \) \( I = \frac{1}{15} \log \left| \frac{z}{z + 3} \right| + C = \frac{1}{15} \log \left| \frac{x^5}{x^5 + 3} \right| + C \)

Question. Find: \( \int \frac{x^2}{x^4 - x^2 - 12} \, dx \)
Answer: Let \( I = \int \frac{x^2}{x^4 - x^2 - 12} \, dx = \int \frac{x^2}{x^4 - 4x^2 + 3x^2 - 12} \, dx \)
\( = \int \frac{x^2}{(x^2 - 4)(x^2 + 3)} \, dx \)
\( = \int \frac{t}{(t - 4)(t + 3)} \, dx = \frac{A}{t - 4} + \frac{B}{t + 3} \quad [ \text{let } x^2 = t ] \)
\( \implies \) \( t = A(t + 3) + B(t - 4) \)
On comparing the coefficient of \( t \) on both sides, we get
\( A + B = 1 \)
\( \implies \) \( 3A - 4B = 0 \)
\( \implies \) \( 3(1 - B) - 4B = 0 \)
\( \implies \) \( 3 - 3B - 4B = 0 \)
\( \implies \) \( 7B = 3 \)
\( \implies \) \( B = \frac{3}{7} \)
If \( B = \frac{3}{7} \), then \( A + \frac{3}{7} = 1 \)
\( \implies \) \( A = 1 - \frac{3}{7} = \frac{4}{7} \)
Now, \( \frac{x^2}{(x^2 - 4)(x^2 + 3)} = \frac{4}{7(x^2 - 4)} + \frac{3}{7(x^2 + 3)} \)
\( \therefore I = \frac{4}{7} \int \frac{1}{x^2 - (2)^2} \, dx + \frac{3}{7} \int \frac{1}{x^2 + (\sqrt{3})^2} \, dx \)
\( = \frac{4}{7 \cdot 2 \cdot 2} \log \left| \frac{x - 2}{x + 2} \right| + \frac{3}{7 \cdot \sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} + C \)
\( = \frac{1}{7} \log \left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{-1} \frac{x}{\sqrt{3}} + C \)

Question. Evaluate: \( \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx \)
Answer: Put \( x^2 = t \), we get
\( \therefore \frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{t}{(t + 4)(t + 9)} \)
Now, \( \frac{t}{(t + 4)(t + 9)} = \frac{A}{t + 4} + \frac{B}{t + 9} = \frac{A(t + 9) + B(t + 4)}{(t + 4)(t + 9)} \)
\( \implies \) \( t = (A + B)t + (9A + 4B) \)
Equating the coefficients, we get
\( A + B = 1, 9A + 4B = 0 \)
Solving above two equations, we get
\( A = -\frac{4}{5}, B = \frac{9}{5} \)
\( \therefore \frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{-4}{5(x^2 + 4)} + \frac{9}{5(x^2 + 9)} \)
\( \therefore I = -\frac{4}{5} \int \frac{dx}{x^2 + 2^2} + \frac{9}{5} \int \frac{dx}{x^2 + 3^2} \)
\( = -\frac{4}{5} \times \frac{1}{2} \tan^{-1} \frac{x}{2} + \frac{9}{5} \times \frac{1}{3} \tan^{-1} \frac{x}{3} + C \)
\( = -\frac{2}{5} \tan^{-1} \frac{x}{2} + \frac{3}{5} \tan^{-1} \frac{x}{3} + C \)

Question. Find: \( \int \frac{(3 \sin \theta - 2) \cos \theta}{5 - \cos^2 \theta - 4 \sin \theta} \, d\theta \)
Answer: We have
\( I = \int \frac{(3 \sin \theta - 2) \cos \theta}{5 - \cos^2 \theta - 4 \sin \theta} \, d\theta \)
Let \( \sin \theta = z \)
\( \implies \) \( \cos \theta \, d\theta = dz \)
\( I = \int \frac{(3z - 2) \, dz}{5 - (1 - z^2) - 4z} = \int \frac{(3z - 2) \, dz}{4 - 4z + z^2} = \int \frac{3z - 2}{(z - 2)^2} \, dz = \int \frac{3z}{(z - 2)^2} \, dz - 2 \int \frac{dz}{(z - 2)^2} \)
Let \( z - 2 = t \)
\( \implies \) \( dz = dt \)
\( = \int \frac{3(t + 2) \, dt}{t^2} - 2 \int \frac{dt}{t^2} = 3 \int \frac{t \, dt}{t^2} + 6 \int \frac{dt}{t^2} - 2 \int \frac{dt}{t^2} = 3 \int \frac{dt}{t} + 4 \int \frac{dt}{t^2} \)
\( = 3 \log | t | + 4 \frac{t^{-2+1}}{-2+1} + C \)
\( = 3 \log | t | - 4 \cdot \frac{1}{t} + C \)
Putting value of \( t \) in terms of \( z \) then \( z \) in terms of \( \theta \), we get
\( = 3 \log | \sin \theta - 2 | - \frac{4}{\sin \theta - 2} + C \)

Question. Find: \( \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx \)
Answer: We have \( I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} \, dx = \int \frac{x^{1/2} \, dx}{\sqrt{a^3 - x^3}} \)
Let \( x^{3/2} = t \)
\( \implies \) \( \frac{3}{2} x^{1/2} \, dx = dt \)
\( \implies \) \( x^{1/2} \, dx = \frac{2}{3} dt \)
\( I = \frac{2}{3} \int \frac{dt}{\sqrt{(a^{3/2})^2 - t^2}} \quad [\because x^{3/2} = t \implies x^3 = t^2] \)
\( = \frac{2}{3} \sin^{-1} \frac{t}{a^{3/2}} + C = \frac{2}{3} \sin^{-1} \sqrt{\frac{x^3}{a^3}} + C \)