CBSE Class 12 Mathematics Integrals Important Questions Set 01

Multiple Choice Questions

Question. \( \int x^2 e^{x^3} dx \) is equal to
(a) \( \frac{1}{3} e^{x^3} + C \)
(b) \( \frac{1}{3} e^{x^4} + C \)
(c) \( \frac{1}{2} e^{x^3} + C \)
(d) \( \frac{1}{2} e^{x^2} + C \)
Answer: (a)

Question. \( \int \frac{e^x(1+x)}{\cos^2(xe^x)} dx \) is equal to
(a) \( \tan(xe^x) + C \)
(b) \( \cot(xe^x) + C \)
(c) \( \cot(e^x) + C \)
(d) \( \tan[e^x(1+x)] + C \)
Answer: (a)

Question. \( \int e^x (\cos x - \sin x) dx \) is equal to
(a) \( e^x \cos x + C \)
(b) \( e^x \sin x + C \)
(c) \( -e^x \cos x + C \)
(d) \( -e^x \sin x + C \)
Answer: (a)

Question. \( \int \frac{dx}{\sin^2 x \cos^2 x} \) is equal to
(a) \( \tan x + \cot x + C \)
(b) \( (\tan x + \cot x)^2 + C \)
(c) \( \tan x - \cot x + C \)
(d) \( (\tan x - \cot x)^2 + C \)
Answer: (c)

Question. If \( \int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx = ax + b \log|4e^x + 5e^{-x}| + C \) then
(a) \( a = -\frac{1}{8}, b = \frac{7}{8} \)
(b) \( a = \frac{1}{8}, b = \frac{7}{8} \)
(c) \( a = -\frac{1}{8}, b = -\frac{7}{8} \)
(d) \( a = \frac{1}{8}, b = -\frac{7}{8} \)
Answer: (a)

Question. \( \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx \) is equal to
(a) \( 2(\sin x + x \cos \theta) + C \)
(b) \( 2(\sin x - x \cos \theta) + C \)
(c) \( 2(\sin x + 2x \cos \theta) + C \)
(d) \( 2(\sin x - 2x \cos \theta) + C \)
Answer: (a)

Question. \( \int \frac{dx}{\sin(x - a) \sin(x - b)} \) is equal to
(a) \( \sin(b - a) \log \left| \frac{\sin(x - b)}{\sin(x - a)} \right| + C \)
(b) \( \csc(b - a) \log \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + C \)
(c) \( \csc(b - a) \log \left| \frac{\sin(x - b)}{\sin(x - a)} \right| + C \)
(d) \( \sin(b - a) \log \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + C \)
Answer: (c)

Question. \( \int \tan^{-1} \sqrt{x} dx \) is equal to
(a) \( (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C \)
(b) \( x \tan^{-1} \sqrt{x} - \sqrt{x} + C \)
(c) \( \sqrt{x} - x \tan^{-1} \sqrt{x} + C \)
(d) \( \sqrt{x} - (x+1) \tan^{-1} \sqrt{x} + C \)
Answer: (a)

Question. \( \int e^x \left( \frac{1 - x}{1 + x^2} \right)^2 dx \) is equal to
(a) \( \frac{e^x}{1 + x^2} + C \)
(b) \( \frac{-e^x}{1 + x^2} + C \)
(c) \( \frac{e^x}{(1 + x^2)^2} + C \)
(d) \( \frac{-e^x}{(1 + x^2)^2} + C \)
Answer: (a)

Question. \( \int \frac{\sin^6 x}{\cos^8 x} dx \) is equal to
(a) \( \frac{\tan^6 x}{5} + C \)
(b) \( \frac{\tan^7 x}{5} + C \)
(c) \( \frac{\tan^7 x}{7} + C \)
(d) none of these
Answer: (c)

Question. If \( \int \frac{x^3}{\sqrt{1 + x^2}} dx = a(1 + x^2)^{3/2} + b\sqrt{1 + x^2} + C \), then
(a) \( a = \frac{1}{3}, b = 1 \)
(b) \( a = -\frac{1}{3}, b = 1 \)
(c) \( a = -\frac{1}{3}, b = -1 \)
(d) \( a = \frac{1}{3}, b = -1 \)
Answer: (d)

Question. The integral \( \int \frac{x^9}{(4x^2 + 1)^6} dx \) is equal to
(a) \( \frac{1}{5x} \left( 4 + \frac{1}{x^2} \right)^{-5} + C \)
(b) \( \frac{1}{5} \left( 4 + \frac{1}{x^2} \right)^{-5} + C \)
(c) \( \frac{1}{10x} (5)^{-5} + C \)
(d) \( \frac{1}{10} \left( \frac{1}{x^2} + 4 \right)^{-5} + C \)
Answer: (d)

Question. \( \int_0^{\pi/8} \tan^2(2x) dx \) is equal to
(a) \( \frac{4 - \pi}{8} \)
(b) \( \frac{4 + \pi}{8} \)
(c) \( \frac{4 - \pi}{4} \)
(d) \( \frac{4 - \pi}{2} \)
Answer: (a)

Question. \( \int_{a+c}^{b+c} f(x) dx \) is equal to
(a) \( \int_a^b f(x - c) dx \)
(b) \( \int_a^b f(x + c) dx \)
(c) \( \int_a^b f(x) dx \)
(d) \( \int_{a-c}^{b-c} f(x) dx \)
Answer: (b)

Question. If \( f \) and \( g \) are continuous functions in \( [0, a] \) satisfying \( f(x) = f(a - x) \) and \( g(x) + g(a - x) = a \), then \( \int_0^a f(x) \cdot g(x) dx \) is equal to
(a) \( \frac{a}{2} \)
(b) \( \frac{a}{2} \int_0^a f(x) dx \)
(c) \( \int_0^a f(x) dx \)
(d) \( a \int_0^a f(x) dx \)
Answer: (b)

Question. \( \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx \) is equal to
(a) \( \log 2 \)
(b) \( 2 \log 2 \)
(c) \( \frac{1}{2} \log 2 \)
(d) \( 4 \log 2 \)
Answer: (b)

Question. \( \int_0^1 \frac{e^t}{1 + t} dt = a \), then \( \int_0^1 \frac{e^t}{(1 + t)^2} dt \) is equal to
(a) \( a - 1 + \frac{e}{2} \)
(b) \( a + 1 - \frac{e}{2} \)
(c) \( a - 1 - \frac{e}{2} \)
(d) \( a + 1 + \frac{e}{2} \)
Answer: (a)

Question. \( \int_{-2}^2 |x \cos \pi x| dx \) is equal to
(a) \( \frac{8}{\pi} \)
(b) \( \frac{4}{\pi} \)
(c) \( \frac{2}{\pi} \)
(d) \( \frac{1}{\pi} \)
Answer: (a)

Question. The integral value of \( \int_0^{\pi/2} \frac{\tan x}{1 + m^2 \tan^2 x} dx \), \( m > 0 \), is
(a) \( \frac{\log m}{(m^2 - 1)} \)
(b) \( \log \left( \frac{m^2 - m}{2} \right) \)
(c) \( \log 3m \)
(d) 0
Answer: (a)

Question. \( \int_{-1}^2 |x| dx \) is equal to
(a) 1
(b) \( \frac{3}{2} \)
(c) 2
(d) \( \frac{5}{2} \)
Answer: (d)

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 \( \sin \theta = z \Rightarrow \cos \theta \, d\theta = dz \)
\( \therefore I = \int \frac{(3z - 2) dz}{5 - (1 - z^2) - 4z} \)
\( = \int \frac{(3z - 2) dz}{5 - 1 + z^2 - 4z} = \int \frac{(3z - 2)}{4 - 4z + z^2} dz = \int \frac{3z - 2}{(z - 2)^2} dz = \int \frac{3z}{(z - 2)^2} dz - 2 \int \frac{dz}{(z - 2)^2} \)
Let \( z - 2 = t \Rightarrow dz = dt \)
\( = \int \frac{3(t + 2) dt}{t^2} - 2 \int \frac{dt}{t^2} = 3 \int \frac{t}{t^2} dt + 6 \int \frac{dt}{t^2} - 2 \int \frac{dt}{t^2} = 3 \int \frac{dt}{t} + 4 \int \frac{dt}{t^2} = 3 \log |t| + 4 \frac{t^{-2+1}}{-2 + 1} + C \)
\( = 3 \log |t| - 4 \cdot \frac{1}{t} + C \)
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 \( I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} dx = \int \frac{x^{1/2} dx}{\sqrt{a^3 - x^3}} = \int \frac{x^{1/2} dx}{\sqrt{(a^{3/2})^2 - (x^{3/2})^2}} \)
Let \( x^{3/2} = t \Rightarrow \frac{3}{2} x^{1/2} dx = dt \Rightarrow x^{1/2} dx = \frac{2}{3} dt \)
\( I = \frac{2}{3} \int \frac{dt}{\sqrt{(a^{3/2})^2 - t^2}} \) \( [\because x^{3/2} = t \Rightarrow x^3 = t^2] \)
\( = \frac{2}{3} \sin^{-1} \left( \frac{t}{a^{3/2}} \right) + C = \frac{2}{3} \sin^{-1} \left( \frac{x^{3/2}}{a^{3/2}} \right) + C \)
\( = \frac{2}{3} \sin^{-1} \sqrt{\frac{x^3}{a^3}} + C \)

Question. Find: \( \int \frac{(2x - 5)e^{2x}}{(2x - 3)^3} dx \)
Answer: We have,
\( \int \frac{(2x - 5)e^{2x}}{(2x - 3)^3} dx = \int e^{2x} \left[ \frac{(2x - 3) - 2}{(2x - 3)^3} \right] dx \)
\( = \int e^3 \cdot e^{2x-3} \left[ \frac{1}{(2x - 3)^2} - \frac{2}{(2x - 3)^3} \right] dx = e^3 \int e^{2x-3} \left[ \frac{1}{(2x - 3)^2} - \frac{2}{(2x - 3)^3} \right] dx \)
Let \( 2x - 3 = t \Rightarrow 2 dx = dt \Rightarrow dx = \frac{dt}{2} \)
\( \Rightarrow I = \frac{e^3}{2} \int e^t \left[ \frac{1}{t^2} - \frac{2}{t^3} \right] dt \Rightarrow I = \frac{e^3}{2} e^t \cdot \frac{1}{t^2} + C \)
Putting \( t = 2x - 3 \)
\( I = \frac{e^3}{2} e^{2x-3} \cdot \frac{1}{(2x - 3)^2} + C \Rightarrow I = \frac{e^{2x}}{2(2x - 3)^2} + C \)

Question. Find : \( \int (2x + 5) \sqrt{10 - 4x - 3x^2} dx \)
Answer: Let, \( I = \int (2x + 5) \sqrt{10 - 4x - 3x^2} dx \)
Let \( (2x + 5) = A \frac{d}{dx}(10 - 4x - 3x^2) + B \)
\( \Rightarrow 2x + 5 = A(-4 - 6x) + B \Rightarrow 2x + 5 = -4A - 6Ax + B \)
Equating, we get
\( -4A + B = 5 \) ...(i) and \( -6A = 2 \) ...(ii)
\( (ii) \Rightarrow A = - \frac{1}{3} \)
Now, from (i) \( \frac{4}{3} + B = 5 \Rightarrow B = 5 - \frac{4}{3} = \frac{11}{3} \)
\( \therefore 2x + 5 = - \frac{1}{3}(-4 - 6x) + \frac{11}{3} \)
Now, \( I = \int \left\{ - \frac{1}{3}(-4 - 6x) + \frac{11}{3} \right\} \sqrt{10 - 4x - 3x^2} dx \)
\( = - \frac{1}{3} \int (-4 - 6x) \sqrt{10 - 4x - 3x^2} dx + \frac{11}{3} \int \sqrt{10 - 4x - 3x^2} dx \)
\( I = -\frac{1}{3} I_1 + \frac{11}{3} I_2 \) ...(iii)
where \( I_1 = \int (-4 - 6x) \sqrt{10 - 4x - 3x^2} dx \) and \( I_2 = \int \sqrt{10 - 4x - 3x^2} dx \)
Now, \( I_1 = \int (-4 - 6x) \sqrt{10 - 4x - 3x^2} dx \)
Let \( 10 - 4x - 3x^2 = z \Rightarrow (-4 - 6x) dx = dz \)
\( \therefore I_1 = \int \sqrt{z} dz = \frac{z^{\frac{1}{2}+1}}{\frac{1}{2} + 1} + C_1 = \frac{2}{3}(10 - 4x - 3x^2)^{3/2} + C_1 \) ...(iv)
Again \( I_2 = \int \sqrt{10 - 4x - 3x^2} dx = \int \sqrt{-3(x^2 + \frac{4}{3}x - \frac{10}{3})} dx \)
\( = \sqrt{3} \int \sqrt{-\{x^2 + 2x \cdot \frac{2}{3} + \frac{4}{9} - \frac{4}{9} - \frac{10}{3}\}} dx = \sqrt{3} \int \sqrt{-\{(x + \frac{2}{3})^2 - \frac{34}{9}\}} dx \)
\( = \sqrt{3} \int \sqrt{\frac{34}{9} - (x + \frac{2}{3})^2} dx = \sqrt{3} \int \sqrt{\left( \frac{\sqrt{34}}{3} \right)^2 - (x + \frac{2}{3})^2} dx \)
\( = \frac{\sqrt{3}}{2} \left( x + \frac{2}{3} \right) \sqrt{10 - 4x - 3x^2} + \frac{\sqrt{3}}{2} \times \frac{34}{9} \sin^{-1} \left( \frac{x + \frac{2}{3}}{\frac{\sqrt{34}}{3}} \right) + C_2 \) ...(v)
Putting the value of \( I_1 \) and \( I_2 \) in (iii), we get
\( I = -\frac{2}{9}(10 - 4x - 3x^2)^{3/2} + \frac{11}{2\sqrt{3}} \left( x + \frac{2}{3} \right) \sqrt{10 - 4x - 3x^2} + \frac{187\sqrt{3}}{27} \sin^{-1} \left( \frac{3}{\sqrt{34}} (x + \frac{2}{3}) \right) + C \)

Question. Evaluate: \( \int_0^1 x \log(1 + 2x) dx \)
Answer: Let \( I = \int_0^1 x \log(1 + 2x) dx \)
Using integration by parts:
\( = \left[ \log(1 + 2x) \cdot \frac{x^2}{2} \right]_0^1 - \int_0^1 \frac{1}{1 + 2x} \cdot 2 \cdot \frac{x^2}{2} dx = \frac{1}{2} [x^2 \log(1 + 2x)]_0^1 - \int_0^1 \frac{x^2}{1 + 2x} dx \)
\( = \frac{1}{2} [1 \log 3 - 0] - \int_0^1 \frac{1}{4} \left[ \frac{4x^2 - 1 + 1}{1 + 2x} \right] dx = \frac{1}{2} \log 3 - \frac{1}{4} \int_0^1 (2x - 1) dx + \frac{1}{2} \int_0^1 \frac{x}{1 + 2x} dx \)
\( = \frac{1}{2} \log 3 - \frac{1}{4} \left[ \frac{2x^2}{2} - x \right]_0^1 + \frac{1}{2} \int_0^1 \frac{1}{2} \frac{(2x + 1 - 1)}{(2x + 1)} dx \)
\( = \frac{1}{2} \log 3 - \frac{1}{4} \left[ x^2 - x \right]_0^1 + \frac{1}{4} \int_0^1 \left( 1 - \frac{1}{2x + 1} \right) dx \)
\( = \frac{1}{2} \log 3 - \frac{1}{4} [1 - 1 - 0] + \frac{1}{4} [x]_0^1 - \frac{1}{8} [\log(1 + 2x)]_0^1 = \frac{1}{2} \log 3 - 0 + \frac{1}{4} - \frac{1}{8} [ \log 3 - \log 1 ] \)
\( = \frac{1}{2} \log 3 + \frac{1}{4} - \frac{1}{8} \log 3 = \frac{3}{8} \log 3 + \frac{1}{4} \)

Question. Evaluate: \( \int_0^\pi \frac{4x \sin x}{1 + \cos^2 x} dx \)
Answer: Let \( I = \int_0^\pi \frac{4x \sin x}{1 + \cos^2 x} dx \) ...(i)
\( = \int_0^\pi \frac{4(\pi - x) \cdot \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx \)
\( I = \int_0^\pi \frac{4(\pi - x) \cdot \sin x}{1 + \cos^2 x} dx \) ...(ii)
Adding (i) and (ii), we get
\( 2I = \int_0^\pi \frac{4(x + \pi - x) \sin x}{1 + \cos^2 x} dx \Rightarrow 2I = 4 \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x} dx \)
\( I = 2\pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx \)
Let \( \cos x = z \Rightarrow - \sin x \, dx = dz \Rightarrow \sin x \, dx = - dz \)
The limits are, \( x = 0 \Rightarrow z = 1 \) and \( x = \pi \Rightarrow z = -1 \)
\( \therefore I = 2\pi \int_1^{-1} \frac{- dz}{1 + z^2} = 2\pi [ \tan^{-1} z ]_{-1}^1 \)
\( = 2\pi [ \tan^{-1} 1 - \tan^{-1}(-1) ] = 2\pi [ \frac{\pi}{4} + \frac{\pi}{4} ] = 2\pi \times \frac{\pi}{2} \)
\( \Rightarrow I = \pi^2 \).

Question. Evaluate: \( \int_{-\pi}^\pi (\cos ax - \sin bx)^2 dx \)
Answer: Here, \( I = \int_{-\pi}^\pi (\cos ax - \sin bx)^2 dx \)
\( \Rightarrow I = \int_{-\pi}^\pi (\cos^2 ax + \sin^2 bx - 2 \cos ax \sin bx) dx \)
\( \Rightarrow I = \int_{-\pi}^\pi \cos^2 ax \, dx + \int_{-\pi}^\pi \sin^2 bx \, dx - \int_{-\pi}^\pi 2 \cos ax \sin bx \, dx \)
\( \Rightarrow I = 2 \int_0^\pi \cos^2 ax \, dx + 2 \int_0^\pi \sin^2 bx \, dx - 0 \) [First two integrands are even function while third is odd function.]
\( \Rightarrow I = 2 \int_0^\pi \frac{(1 + \cos 2ax)}{2} dx + 2 \int_0^\pi \frac{(1 - \cos 2bx)}{2} dx \)
\( \Rightarrow I = \int_0^\pi dx + \int_0^\pi \cos 2ax \, dx + \int_0^\pi dx - \int_0^\pi \cos 2bx \, dx = \int_0^\pi 2 \, dx + \int_0^\pi \cos 2ax \, dx - \int_0^\pi \cos 2bx \, dx \)
\( \Rightarrow I = 2[x]_0^\pi + \left[ \frac{\sin 2ax}{2a} \right]_0^\pi - \left[ \frac{\sin 2bx}{2b} \right]_0^\pi \)
\( \Rightarrow I = 2\pi + \frac{\sin 2a\pi}{2a} - \frac{\sin 2b\pi}{2b} \)

Question. Evaluate: \( \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx \)
Answer: Let \( I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx \)
\( I = \int_0^\pi \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx \)
\( = \int_0^\pi \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx = \pi \int_0^\pi \frac{\sin x \, dx}{1 + \cos^2 x} - I \)
or \( 2I = \pi \int_0^\pi \frac{\sin x \, dx}{1 + \cos^2 x} \) or \( I = \frac{\pi}{2} \int_0^\pi \frac{\sin x \, dx}{1 + \cos^2 x} \)
Put \( \cos x = t \) so that \( - \sin x \, dx = dt \).
The limits are, when \( x = 0, t = 1 \) and \( x = \pi, t = -1 \), we get
\( I = \frac{-\pi}{2} \int_1^{-1} \frac{dt}{1 + t^2} = \pi \int_0^1 \frac{dt}{1 + t^2} \) [\( \because \int_{-a}^a f(x) dx = - \int_a^{-a} f(x) dx \) and \( \int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx \)]
\( = \pi [ \tan^{-1} t ]_0^1 = \pi [ \tan^{-1} 1 - \tan^{-1} 0 ] = \pi [ \frac{\pi}{4} - 0 ] = \frac{\pi^2}{4} \)

Question. Find: \( \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \)
Answer: Let \( I = \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2 \cdot 2 \sin x \cos x}} \)
\( = \frac{1}{2} \int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{\frac{\sin x}{\cos x} \cdot \cos^2 x}} = \frac{1}{2} \int_0^{\pi/4} \frac{dx}{\cos^4 x \sqrt{\tan x}} \)
\( = \frac{1}{2} \int_0^{\pi/4} \frac{\sec^4 x \, dx}{\sqrt{\tan x}} = \frac{1}{2} \int_0^{\pi/4} \frac{\sec^2 x \cdot \sec^2 x \, dx}{\sqrt{\tan x}} \)
Let \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \), \( x = 0 \Rightarrow t = 0 \) and \( x = \frac{\pi}{4} \Rightarrow t = 1 \)
\( \therefore I = \frac{1}{2} \int_0^1 \frac{(1 + t^2) dt}{\sqrt{t}} = \frac{1}{2} \int_0^1 (t^{-1/2} + t^{3/2}) dt = \frac{1}{2} \left[ \frac{t^{-1/2+1}}{-1/2+1} \right]_0^1 + \frac{1}{2} \left[ \frac{t^{3/2+1}}{3/2+1} \right]_0^1 \)
\( = \frac{1}{2} \times 2 [ \sqrt{t} ]_0^1 + \frac{1}{2} \times \frac{2}{5} [ t^{5/2} ]_0^1 = 1 + \frac{1}{5} = \frac{6}{5} \)

Question. Evaluate: \( \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1 + e^x} dx \)
Answer: Let \( I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1 + e^x} dx \)
\( = \int_{-\pi/2}^0 \frac{\cos x}{1 + e^x} dx + \int_0^{\pi/2} \frac{\cos x}{1 + e^x} dx \)
In 1st Integrand, Let \( x = -t \Rightarrow dx = -dt \)
\( x = -\pi/2 \Rightarrow t = \pi/2 \), \( x = 0 \Rightarrow t = 0 \)
\( = \int_{\pi/2}^0 \frac{\cos t}{1 + e^{-t}} (-dt) + \int_0^{\pi/2} \frac{\cos x}{1 + e^x} dx \)
\( = \int_0^{\pi/2} \frac{\cos t}{1 + \frac{1}{e^t}} dt + \int_0^{\pi/2} \frac{\cos x}{1 + e^x} dx = \int_0^{\pi/2} \frac{e^t \cdot \cos t}{1 + e^t} dt + \int_0^{\pi/2} \frac{\cos x}{1 + e^x} dx \)
\( = \int_0^{\pi/2} \frac{e^x \cos x}{1 + e^x} dx + \int_0^{\pi/2} \frac{\cos x}{1 + e^x} dx \) [By property \( \int_a^b f(x) dx = \int_a^b f(t) dt \)]
\( = \int_0^{\pi/2} \frac{(e^x + 1) \cdot \cos x}{1 + e^x} dx = \int_0^{\pi/2} \cos x \, dx = [ \sin x ]_0^{\pi/2} = \sin \pi/2 - \sin 0 = 1 \).

Question. Evaluate: \( \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} \)
Answer: Let \( I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} \)
\( = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan(\frac{\pi}{6} + \frac{\pi}{3} - x)}} \) [By using property \( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \)] ...(i)
\( = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan(\frac{\pi}{2} - x)}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \frac{1}{\sqrt{\tan x}}} \)
\( = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} dx \) ...(ii)
Adding (i) and (ii), we get
\( 2I = \int_{\pi/6}^{\pi/3} \frac{(1 + \sqrt{\tan x})}{(1 + \sqrt{\tan x})} dx \)
\( = \int_{\pi/6}^{\pi/3} dx = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)
\( 2I = \frac{\pi}{6} \quad \text{or} \quad I = \frac{\pi}{12} \)

Question. Evaluate: \( \int_1^3 [|x-1| + |x-2| + |x-3|] dx \)
Answer: Let \( I = \int_1^3 [|x-1| + |x-2| + |x-3|] dx = \int_1^3 |x-1| dx + \int_1^3 |x-2| dx + \int_1^3 |x-3| dx \)
\( = \int_1^3 |x-1| dx + \int_1^2 |x-2| dx + \int_2^3 |x-2| dx + \int_1^3 |x-3| dx \)
[By property of definite integral]
\( = \int_1^3 (x-1) dx + \int_1^2 -(x-2) dx + \int_2^3 (x-2) dx + \int_1^3 -(x-3) dx \)
\( \begin{cases} x-1 \ge 0, & \text{if } 1 \le x \le 3 \\ x-2 \le 0, & \text{if } 1 \le x \le 2 \\ x-2 \ge 0, & \text{if } 2 \le x \le 3 \\ x-3 \le 0, & \text{if } 1 \le x \le 3 \end{cases} \)
\( = \left[ \frac{(x-1)^2}{2} \right]_1^3 - \left[ \frac{(x-2)^2}{2} \right]_1^2 + \left[ \frac{(x-2)^2}{2} \right]_2^3 - \left[ \frac{(x-3)^2}{2} \right]_1^3 \)
\( = \left( \frac{4}{2} - 0 \right) - \left( 0 - \frac{1}{2} \right) + \left( \frac{1}{2} - 0 \right) - \left( 0 - \frac{4}{2} \right) = 2 + \frac{1}{2} + \frac{1}{2} + 2 = 5 \)

Question. Evaluate: \( \int_0^\pi \frac{x \tan x}{\sec x \csc x} dx \)
Answer: Let \( I = \int_0^\pi \frac{x \tan x}{\sec x \csc x} dx = \int_0^\pi x \cdot \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} \cdot \frac{1}{\sin x}} dx \)
\( I = \int_0^\pi x \sin^2 x \, dx = \int_0^\pi (\pi - x) \sin^2 (\pi - x) dx \) [\( \because \int_0^a f(x) dx = \int_0^a f(a - x) dx \)]
\( I = \int_0^\pi \pi \sin^2 x \, dx - \int_0^\pi x \sin^2 x \, dx \Rightarrow 2I = \pi \int_0^\pi \sin^2 x \, dx \)
\( = \frac{\pi}{2} \int_0^\pi (1 - \cos 2x) dx = \frac{\pi}{2} \int_0^\pi dx - \frac{\pi}{2} \int_0^\pi \cos 2x \, dx = \frac{\pi}{2} [x]_0^\pi - \frac{\pi}{2} \left[ \frac{\sin 2x}{2} \right]_0^\pi \)
\( 2I = \frac{\pi}{2} (\pi - 0) - \frac{\pi}{4} (\sin 2\pi - \sin 0) \)
\( 2I = \frac{\pi^2}{2} - 0 \Rightarrow I = \frac{\pi^2}{4} \)

Question. Evaluate \( \int \frac{dx}{\sin(x - a) \cos(x - b)} \).
Answer: Let \( I = \int \frac{dx}{\sin(x - a) \cos(x - b)} = \frac{1}{\cos(a - b)} \int \frac{\cos(a - b) dx}{\sin(x - a) \cos(x - b)} \)
\( = \frac{1}{\cos(a - b)} \int \frac{\cos((x - b) - (x - a)) dx}{\sin(x - a) \cos(x - b)} \)
\( = \frac{1}{\cos(a - b)} \int \frac{\cos(x - b) \cos(x - a) + \sin(x - b) \sin(x - a)}{\sin(x - a) \cos(x - b)} dx \)
\( = \frac{1}{\cos(a - b)} \int \left\{ \frac{\cos(x - a)}{\sin(x - a)} + \frac{\sin(x - b)}{\cos(x - b)} \right\} dx = \frac{1}{\cos(a - b)} \left[ \int \frac{\cos(x - a)}{\sin(x - a)} dx + \int \frac{\sin(x - b)}{\cos(x - b)} dx \right] \)
\( = \frac{1}{\cos(a - b)} [ \log |\sin(x - a)| - \log |\cos(x - b)| ] + C \)
\( = \frac{1}{\cos(a - b)} \log \left| \frac{\sin(x - a)}{\cos(x - b)} \right| + C \)

Question. Evaluate: \( \int_0^1 \cot^{-1}(1 - x + x^2) dx \)
Answer: Let \( I = \int_0^1 \cot^{-1}(1 - x + x^2) dx \)
\( = \int_0^1 \tan^{-1} \left( \frac{1}{1 - x + x^2} \right) dx \) [\( \because \cot^{-1} x = \tan^{-1} \frac{1}{x} \)]
\( = \int_0^1 \tan^{-1} \frac{x + (1 - x)}{1 - x(1 - x)} dx \)
\( = \int_0^1 \{ \tan^{-1} x + \tan^{-1}(1 - x) \} dx \) [\( \because \tan^{-1}(x + y) = \tan^{-1} \frac{x + y}{1 - xy} \)]
\( = \int_0^1 \tan^{-1} x \, dx + \int_0^1 \tan^{-1}(1 - x) dx \)
\( = \int_0^1 \tan^{-1} x \, dx + \int_0^1 \tan^{-1}(1 - (1 - x)) dx \) [\( \because \int_0^a f(x) dx = \int_0^a f(a - x) dx \)]
\( = \int_0^1 \tan^{-1} x \, dx + \int_0^1 \tan^{-1} x \, dx = 2 \int_0^1 \tan^{-1} x \, dx = 2 \int_0^1 1 \cdot \tan^{-1} x \, dx \)
\( = 2 [ x \cdot \tan^{-1} x ]_0^1 - 2 \int_0^1 \frac{x}{1 + x^2} dx = 2 \times \frac{\pi}{4} - \int_0^1 \frac{2x \, dx}{1 + x^2} = \frac{\pi}{2} - [ \log |1 + x^2| ]_0^1 \)
\( = \frac{\pi}{2} - [ \log 2 - \log 1 ] = \frac{\pi}{2} - \log 2 \)

Question. Evaluate: \( \int_0^1 x^2(1 - x)^n dx \)
Answer: Let \( I = \int_0^1 x^2(1 - x)^n dx \)
\( I = \int_0^1 (1 - x)^2 [1 - (1 - x)]^n dx \) [\( \because \int_0^a f(x) dx = \int_0^a f(a - x) dx \)]
\( = \int_0^1 (1 - 2x + x^2) x^n dx = \int_0^1 (x^n - 2x^{n+1} + x^{n+2}) dx \)
\( = \left[ \frac{x^{n+1}}{n+1} - 2 \cdot \frac{x^{n+2}}{n+2} + \frac{x^{n+3}}{n+3} \right]_0^1 = \frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3} \)
\( = \frac{(n+2)(n+3) - 2(n+1)(n+3) + (n+1)(n+2)}{(n+1)(n+2)(n+3)} \)
\( = \frac{n^2 + 5n + 6 - 2n^2 - 8n - 6 + n^2 + 3n + 2}{(n+1)(n+2)(n+3)} = \frac{2}{(n+1)(n+2)(n+3)} \)

Question. Evaluate: \( \int_1^3 (2x^2 + 5x) dx \) as a limit of a sum.
Answer: Let \( f(x) = 2x^2 + 5x \)
Here \( a = 1, b = 3 \therefore h = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n} \Rightarrow nh = 2 \)
Also, \( n \to \infty \iff h \to 0 \).
\( \int_a^b f(x) dx = \lim_{h \to 0} h [ f(a) + f(a + h) + f(a + 2h) + \dots + f(a + (n-1)h) ] \)
\( \int_1^3 (2x^2 + 5x) dx = \lim_{h \to 0} h [ f(1) + f(1 + h) + f(1 + 2h) + \dots + f(1 + (n-1)h) ] \)
\( = \lim_{h \to 0} h [ \{ 2 \cdot 1^2 + 5 \cdot 1 \} + \{ 2(1 + h)^2 + 5(1 + h) \} + \dots + \{ 2(1 + (n-1)h)^2 + 5(1 + (n-1)h) \} ] \)
\( = \lim_{h \to 0} h [ (2 + 5) + \{ 2 + 4h + 2h^2 + 5 + 5h \} + \dots + \{ 2 + 4(n-1)h + 2(n-1)^2h^2 + 5 + 5(n-1)h \} ] \)
\( = \lim_{h \to 0} h [ 7 + \{ 7 + 9h + 2h^2 \} + \dots + \{ 7 + 9(n-1)h + 2(n-1)^2h^2 \} ] \)
\( = \lim_{h \to 0} h [ 7n + 9h \{ 1 + 2 + \dots + (n-1) \} + 2h^2 \{ 1^2 + 2^2 + \dots + (n-1)^2 \} ] \)
\( = \lim_{h \to 0} \left[ 7nh + 9h^2 \frac{(n-1) \cdot n}{2} + 2h^3 \frac{(n-1) \cdot n (2n-1)}{6} \right] \)
\( = \lim_{h \to 0} \left[ 7(nh) + \frac{9(nh)^2 \cdot (1 - \frac{1}{n})}{2} + \frac{2(nh)^3 (1 - \frac{1}{n}) \cdot (2 - \frac{1}{n})}{6} \right] \)
\( = \lim_{n \to \infty} \left[ 14 + \frac{36(1 - \frac{1}{n})}{2} + \frac{16(1 - \frac{1}{n}) \cdot (2 - \frac{1}{n})}{6} \right] \) [\( \because nh = 2 \)]
\( = \lim_{n \to \infty} \left[ 14 + 18(1 - \frac{1}{n}) + \frac{8}{3}(1 - \frac{1}{n}) \cdot (2 - \frac{1}{n}) \right] \)
\( = 14 + 18 + \frac{8}{3} \times 1 \times 2 = 32 + \frac{16}{3} = \frac{96 + 16}{3} = \frac{112}{3} \)

Question. Evaluate : \( \int_{-1}^2 |x^3 - x| dx \)
Answer: If \( x^3 - x = 0 \)
\( \Rightarrow x(x^2 - 1) = 0 \Rightarrow x = 0 \) or \( x^2 = 1 \)
\( \Rightarrow x = 0 \) or \( x = \pm 1 \Rightarrow x = 0, -1, 1 \)
Hence [–1, 2] is divided into three sub intervals [–1, 0], [0, 1] and [1, 2] such that
\( x^3 - x \ge 0 \) on [–1, 0]
\( x^3 - x \le 0 \) on [0, 1]
and \( x^3 - x \ge 0 \) on [1, 2]
Now, \( \int_{-1}^2 |x^3 - x| dx = \int_{-1}^0 |x^3 - x| dx + \int_0^1 |x^3 - x| dx + \int_1^2 |x^3 - x| dx \)
\( = \int_{-1}^0 (x^3 - x) dx + \int_0^1 -(x^3 - x) dx + \int_1^2 (x^3 - x) dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^0 - \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_0^1 + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_1^2 \)
\( = \{ 0 - (\frac{1}{4} - \frac{1}{2}) \} - \{ (\frac{1}{4} - \frac{1}{2}) - 0 \} + \{ (4 - 2) - (\frac{1}{4} - \frac{1}{2}) \} \)
\( = - \frac{1}{4} + \frac{1}{2} - \frac{1}{4} + \frac{1}{2} + 2 - \frac{1}{4} + \frac{1}{2} = \frac{3}{2} - \frac{3}{4} + 2 = \frac{11}{4} \)

Question. Evaluate \( \int_0^\pi e^{2x} \cdot \sin(\frac{\pi}{4} + x) dx \).
Answer: We have \( I = \int_0^\pi e^{2x} \cdot \sin(\frac{\pi}{4} + x) dx \)
Integrating by part, we get
\( I = \left[ \sin(\frac{\pi}{4} + x) \cdot \frac{e^{2x}}{2} \right]_0^\pi - \int_0^\pi \cos(\frac{\pi}{4} + x) \cdot \frac{e^{2x}}{2} dx \)
\( = \frac{1}{2} \left[ \sin \frac{5\pi}{4} \cdot e^{2\pi} - \sin \frac{\pi}{4} \right] - \frac{1}{2} \int_0^\pi e^{2x} \cdot \cos(\frac{\pi}{4} + x) dx \)
\( = \frac{1}{2} \left[ -\frac{e^{2\pi}}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right] - \frac{1}{2} \left[ \{ \cos(\frac{\pi}{4} + x) \cdot \frac{e^{2x}}{2} \}_0^\pi + \int_0^\pi \sin(\frac{\pi}{4} + x) \cdot \frac{e^{2x}}{2} dx \right] \)
\( = - \frac{e^{2\pi} + 1}{2\sqrt{2}} - \frac{1}{2} \left[ \cos \frac{5\pi}{4} \cdot \frac{e^{2\pi}}{2} - \cos \frac{\pi}{4} \cdot \frac{1}{2} \right] - \frac{1}{4} \int_0^\pi e^{2x} \cdot \sin(\frac{\pi}{4} + x) dx \)
\( I = - \frac{e^{2\pi} + 1}{2\sqrt{2}} - \frac{1}{4} e^{2\pi} \cdot \cos \frac{5\pi}{4} + \frac{1}{4\sqrt{2}} - \frac{1}{4} I \)
\( \frac{5I}{4} = - \frac{e^{2\pi} + 1}{2\sqrt{2}} + \frac{e^{2\pi}}{4\sqrt{2}} + \frac{1}{4\sqrt{2}} = - \frac{e^{2\pi} + 1}{2\sqrt{2}} + \frac{e^{2\pi} + 1}{4\sqrt{2}} = \frac{e^{2\pi} + 1}{4\sqrt{2}} (-2 + 1) = - \frac{e^{2\pi} + 1}{4\sqrt{2}} \)
\( I = - \frac{e^{2\pi} + 1}{5\sqrt{2}} \)

Question. Evaluate : \( \int_{-2}^2 \frac{x^2}{1 + 5^x} dx \)
Answer: Let \( I = \int_{-2}^2 \frac{x^2}{1 + 5^x} dx \) ...(i)
\( = \int_{-2}^2 \frac{(2 + (-2) - x)^2}{1 + 5^{(2 + (-2) - x)}} dx \) \( [ \int_a^b f(x) dx = \int_a^b f(a + b - x) dx ] \)
\( = \int_{-2}^2 \frac{(-x)^2}{1 + 5^{-x}} dx = \int_{-2}^2 \frac{x^2}{1 + \frac{1}{5^x}} dx \)
\( I = \int_{-2}^2 \frac{5^x x^2}{1 + 5^x} dx \) ...(ii)
Adding (i) and (ii), we get
\( 2I = \int_{-2}^2 \frac{(1 + 5^x) x^2}{1 + 5^x} dx = \int_{-2}^2 x^2 dx = \left[ \frac{x^3}{3} \right]_{-2}^2 \)
\( \Rightarrow 2I = \frac{1}{3} [8 - (-8)] \Rightarrow I = \frac{16}{3 \times 2} = \frac{8}{3} \)

Question. Find : \( \int [ \log(\log x) + \frac{1}{(\log x)^2} ] dx \)
Answer: Let \( I = \int [ \log(\log x) + \frac{1}{(\log x)^2} ] dx \)
Let \( \log x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt \)
\( \therefore I = \int \{ \log t + \frac{1}{t^2} \} e^t dt \)
\( = \int \{ \log t + \frac{1}{t} - \frac{1}{t} + \frac{1}{t^2} \} e^t dt = \int ( \log t + \frac{1}{t} ) e^t + ( - \frac{1}{t} + \frac{1}{t^2} ) e^t dt \)
\( = e^t \cdot \log t - \frac{1}{t} \cdot e^t + C \) \( [\because \int f(x) + f'(x) e^x dx = f(x) e^x + C] \)
\( = e^{\log x} \log(\log x) - \frac{1}{\log x} e^{\log x} + C \) [Put \( t = \log x \)]
\( = x. \log(\log x) - \frac{x}{\log x} + C \)

Question. Find: \( \int 5^{5^{5^x}} \cdot 5^{5^x} \cdot 5^x dx \)
Answer: Let \( I = \int 5^{5^{5^x}} \cdot 5^{5^x} \cdot 5^x dx \)
Putting \( 5^x = t \Rightarrow 5^x \log 5 \, dx = dt \) or \( 5^x \, dx = \frac{dt}{(\log 5)} \)
Therefore, \( I = \int 5^{5^t} \cdot 5^t \cdot \frac{dt}{(\log 5)} = \frac{1}{(\log 5)} \int 5^{5^t} \cdot 5^t dt \)
Again, putting \( 5^t = u, 5^t dt = \frac{du}{(\log 5)} \)
Therefore, \( I = \frac{1}{(\log 5)} \int 5^u \cdot \frac{du}{(\log 5)} = \frac{1}{(\log 5)^2} \int 5^u du = \frac{5^u}{(\log 5)^2 \cdot (\log 5)} + C \)
\( = \frac{5^u}{(\log 5)^3} + C = \frac{5^{5^t}}{(\log 5)^3} + C = \frac{5^{5^{5^x}}}{(\log 5)^3} + C \)