Definite Integration JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Evalaute
(i) \( \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2x + 2} \)
(ii) \( \int_{\sqrt{2}}^{\infty} \frac{dx}{x\sqrt{x^2 - 1}} \)
(iii) \( \int_{0}^{4} \frac{x^2}{1 + x} dx \)
Answer: (i) \( I = \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2x + 2} = \int_{-\infty}^{\infty} \frac{dx}{1 + (x + 1)^2} = \tan^{-1} (x + 1) \big|_{-\infty}^{\infty} = \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) = \pi \)
(ii) \( I = \int_{\sqrt{2}}^{\infty} \frac{dx}{x\sqrt{x^2 - 1}} = \sec^{-1} x \big|_{\sqrt{2}}^{\infty} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \)
(iii) \( I = \int_{0}^{4} \frac{x^2 + 1 - 1}{x + 1} dx = \int_{0}^{4} (x - 1)dx + \int_{0}^{4} \frac{dx}{1 + x} \)
\( = \left[ \frac{x^2}{2} - x \right]_{0}^{4} + \left[ \ln(1+x) \right]_{0}^{4} = \frac{16}{2} - 4 + \ln 5 = 4 + \ln 5 \)

Question. Let \( f(x) = \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) \), then show that
\( \int_{a}^{b} f(x) dx = \int_{b}^{a} \ln \left( \frac{1 + \sin x}{1 - \sin x} \right) dx \)
Answer: \( f(x) = \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) \)
\( f(-x) = \ln \left( \frac{1 + \sin x}{1 - \sin x} \right) = -\ln \left( \frac{1 - \sin x}{1 + \sin x} \right) = -f(x) \)
Odd function
\( \int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx \)
\( = - \int_{b}^{a} \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) dx = \int_{b}^{a} \ln \left( \frac{1 + \sin x}{1 - \sin x} \right) dx \)

Question. Evaluate
(i) \( \int_{0}^{2} [x^2] dx \)
(ii) \( \int_{-1}^{1} [\cos^{-1} x] dx \)
Answer: (i) \( I = \int_{0}^{2} [x^2] dx \)
Put \( x^2 = t \)
\( x dx = \frac{dt}{2} \)
\( dx = \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int_{0}^{4} \frac{[t]}{\sqrt{t}} dt \)
\( = \frac{1}{2} \left[ \int_{0}^{1} \frac{0}{\sqrt{t}} dt + \int_{1}^{2} \frac{1}{\sqrt{t}} dt + \int_{2}^{3} \frac{2}{\sqrt{t}} dt + \int_{3}^{4} \frac{3}{\sqrt{t}} dt \right] \)
\( = \frac{1}{2} [ (2 t^{1/2})_{1}^{2} + 4 (\sqrt{t})_{2}^{3} + 6 (\sqrt{t})_{3}^{4} ] \)
\( = \frac{1}{2} [ 2\sqrt{2} - 2 + 4\sqrt{3} - 4\sqrt{2} + 6 \times 2 - 6\sqrt{3} ] = 5 - \sqrt{3} - \sqrt{2} \)
 (ii) \( I = \int_{-1}^{1} [\cos^{-1} x] dx \)
\( \cos^{-1} x = t \)
\( x = \cos t \)
\( dx = -\sin t dt \)
\( = - \int_{\pi}^{0} [t] \sin t dt = \int_{0}^{\pi} [t] \sin t dt \)
\( = \int_{0}^{1} 0.\sin t dt + \int_{1}^{2} 1.\sin t dt + \int_{2}^{3} 2.\sin t dt + \int_{3}^{\pi} 3.\sin t dt \)
\( = - [\cos t]_{1}^{2} + 2 [\cos t]_{2}^{3} + 3 [\cos t]_{3}^{\pi} \)
\( = - [\cos 2 - \cos 1 + 2 \cos 3 - 2 \cos 2 - 3 - 3 \cos 3] \)
\( = - [-\cos 2 - \cos 1 - \cos 3 - 3] \)
\( = [3 + \cos 1 + \cos 2 + \cos 3] \)

Question. Evalaute
(i) \( \int_{-1}^{1} e^{|x|} dx \)
(ii) \( \int_{-\pi/4}^{\pi/4} |\sin x| dx \)
(iii) \( \int_{-5}^{5} |x + 2| dx \)
(iv) \( \int_{-\pi/4}^{\pi/4} \frac{x + \pi/4}{2 - \cos 2x} dx \)
Answer: (i) \( I = \int_{-1}^{1} e^{|x|} dx \)
\( = \int_{-1}^{0} e^{-x} dx + \int_{0}^{1} e^x dx = -e^{-x} \Big|_{-1}^{0} + e^x \Big|_{0}^{1} = -[1 - e] + [e - 1] = 2e - 2 \)
 (ii) \( I = \int_{-\pi/4}^{\pi/4} |\sin x| dx \)
\( I = \int_{-\pi/4}^{0} -\sin x dx + \int_{0}^{\pi/4} \sin x dx \)
\( = \cos x \big|_{-\pi/4}^{0} - \cos x \big|_{0}^{\pi/4} = \left( 1 - \frac{1}{\sqrt{2}} \right) - \left( \frac{1}{\sqrt{2}} - 1 \right) = 2 - \sqrt{2} \)
 (iii) \( I = \int_{-5}^{5} |x + 2| dx \)
\( = \int_{-5}^{-2} |x + 2| dx + \int_{-2}^{5} |x + 2| dx \)
\( = -\int_{-5}^{-2} (x + 2) dx + \int_{-2}^{5} (x + 2) dx \)
\( = - \left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5} \)
\( = - \left[ 2 - 4 - \frac{25}{2} + 10 \right] + \left[ \frac{25}{2} + 10 - 2 + 4 \right] = -8 + \frac{25}{2} + \frac{25}{2} + 12 = 4 + 25 = 29 \)
 (iv) \( I = \int_{-\pi/4}^{\pi/4} \frac{x + \pi/4}{2 - \cos 2x} dx \)
\( = \int_{-\pi/4}^{\pi/4} \underbrace{ \frac{x dx}{2 - \cos 2x} }_{\text{Odd Function}} + \frac{\pi}{4} \int_{-\pi/4}^{\pi/4} \underbrace{ \frac{dx}{2 - \cos 2x} }_{\text{Even function}} \)
\( = 0 + \frac{\pi}{2} \int_{0}^{\pi/4} \frac{dx}{2 - \cos 2x} = \frac{\pi}{2} \int_{0}^{\pi/4} \frac{1 + \tan^2 x}{2(1 + \tan^2 x) - (1 - \tan^2 x)} dx \)
\( = \frac{\pi}{2} \int_{0}^{\pi/4} \frac{\sec^2 x dx}{1 + 3 \tan^2 x} \)
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( = \frac{\pi}{2} \int_{0}^{1} \frac{dt}{1 + 3t^2} = \frac{\pi}{6} \int_{0}^{1} \frac{dt}{\frac{1}{3} + t^2} \)
\( = \frac{\pi}{6} \sqrt{3} \tan^{-1} (\sqrt{3} t) \big|_{0}^{1} = \frac{\sqrt{3}\pi}{6} [\tan^{-1} \sqrt{3} - 0] = \frac{\sqrt{3}\pi}{6} \times \frac{\pi}{3} = \frac{\sqrt{3}\pi^2}{18} = \frac{\pi^2}{6\sqrt{3}} \)

Question. Evaluate
(i) \( \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1 + x^2} \right) dx \)
(ii) \( \int_{0}^{1} \frac{x \tan^{-1} x}{(1 + x^2)^{3/2}} dx \)
(iii) \( \int_{0}^{1} x^2 \sin^{-1} x dx \)
(iv) \( \int_{0}^{\sqrt{3}} \tan^{-1} \left( \frac{2x}{1 - x^2} \right) dx \)
Answer: (i) \( I = \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1 + x^2} \right) dx \)
Put \( x = \tan \theta \)
\( dx = \sec^2 \theta d\theta \)
\( = \int_{0}^{\pi/4} \sin^{-1}(\sin 2\theta) \sec^2 \theta d\theta \quad \left(0 < \theta < \frac{\pi}{4}\right) \)
\( = \int_{0}^{\pi/4} 2\theta \sec^2 \theta d\theta \quad \left(0 < 2\theta < \frac{\pi}{2}\right) \)
\( = 2 \int_{0}^{\pi/4} \theta \sec^2 \theta d\theta \)
\( = 2 \left[ [\theta \tan \theta]_{0}^{\pi/4} - \int_{0}^{\pi/4} \tan \theta d\theta \right] \)
\( = 2 \left[ \frac{\pi}{4} - \ln \sqrt{2} \right] = \frac{\pi}{2} - 2\ln \sqrt{2} = \frac{\pi}{2} - \ln 2 \)
 (ii) \( I = \int_{0}^{1} \frac{x \tan^{-1} x}{(1 + x^2)^{3/2}} dx \)
Put \( x = \tan \theta \Rightarrow dx = \sec^2 \theta d\theta \)
\( = \int_{0}^{\pi/4} \frac{\tan \theta \cdot \sec^2 \theta d\theta}{\sec^3 \theta} = \int_{0}^{\pi/4} \theta \sin \theta d\theta \)
\( = [-\theta \cos \theta]_{0}^{\pi/4} + \int_{0}^{\pi/4} \sin \theta d\theta \)
\( = -\frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{4 - \pi}{4\sqrt{2}} \)
 (iii) \( I = \int_{0}^{1} x^2 \sin^{-1} x dx \)
Put \( \sin^{-1} x = \theta \)
\( x = \sin \theta \)
\( dx = \cos \theta d\theta \)
\( = \int_{0}^{\pi/2} \theta \sin^2 \theta \cos \theta d\theta \)
\( = \left[ \theta \cdot \frac{\sin^3 \theta}{3} \right]_{0}^{\pi/2} - \frac{1}{3} \int_{0}^{\pi/2} 1 \cdot \sin^3 \theta d\theta \)
\( = \frac{\pi}{2} \cdot \frac{1}{3} - \frac{1}{3} \int_{0}^{\pi/2} \sin \theta (1 - \cos^2 \theta) d\theta \)
\( = \frac{\pi}{6} - \frac{1}{3} \left[ -\cos \theta + \frac{\cos^3 \theta}{3} \right]_{0}^{\pi/2} = \frac{\pi}{6} - \frac{1}{3} \left[ 0 - \left( -1 + \frac{1}{3} \right) \right] = \frac{\pi}{6} - \frac{2}{9} \)
 (iv) \( I = \int_{0}^{\sqrt{3}} \tan^{-1} \left( \frac{2x}{1 - x^2} \right) dx \)
Put \( x = \tan \theta \Rightarrow dx = \sec^2 \theta d\theta \)
\( = \int_{0}^{\pi/3} \tan^{-1}(\tan 2\theta) \sec^2 \theta d\theta \)
\( = \int_{0}^{\pi/4} 2\theta \sec^2 \theta d\theta + \int_{\pi/4}^{\pi/3} (2\theta - \pi) \sec^2 \theta d\theta \)
\( = \pi \left( 1 - \frac{1}{\sqrt{3}} \right) - \ln 4 \)

Question. Evaluate
(i) \( \int_{0}^{\pi/2} \frac{\sin 2\theta d\theta}{\sin^4 \theta + \cos^4 \theta} \)
(ii) \( \int_{0}^{\pi/2} \sqrt{\cos \theta} \sin^3 \theta d\theta \)
(iii) \( \int_{0}^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
Answer: (i) \( I = \int_{0}^{\pi/2} \frac{\sin 2\theta d\theta}{\sin^4 \theta + \cos^4 \theta} \)
\( I = \int_{0}^{\pi/2} \frac{2\sin \theta \cos \theta d\theta}{\sin^4 \theta + \cos^4 \theta} = \int_{0}^{\pi/2} \frac{2\tan \theta \sec^2 \theta}{1 + \tan^4 \theta} d\theta \)
Put \( \tan^2 \theta = t \)
\( 2\tan \theta \sec^2 \theta d\theta = dt \)
\( = \int_{0}^{\infty} \frac{dt}{1 + t^2} = \tan^{-1} (t) \big|_{0}^{\infty} = \frac{\pi}{2} \)
 (ii) \( I = \int_{0}^{\pi/2} \sqrt{\cos \theta} \sin^3 \theta d\theta \)
Put \( \cos \theta = t^2 \)
\( \sin \theta d\theta = -2t dt \)
\( = -2 \int_{1}^{0} t (\sin^2 \theta) t dt = -2 \int_{1}^{0} t^2 (1 - t^4) dt = 2 \int_{0}^{1} (t^2 - t^6) dt \)
\( = 2 \left[ \frac{t^3}{3} - \frac{t^7}{7} \right]_{0}^{1} = 2 \left[ \frac{1}{3} - \frac{1}{7} \right] = \frac{8}{21} \)
 (iii) \( I = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
\( = \int_{0}^{\pi/4} \frac{\sin x + \cos x}{9 + 16[1 - (\sin x - \cos x)^2]} dx \)
Put \( \sin x - \cos x = t \)
\( (\cos x + \sin x) dx = dt \)
\( = \int_{-1}^{0} \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{1}{16} \int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2} \)
\( = \frac{1}{16} \cdot \frac{1}{2 \times \frac{5}{4}} \left[ \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0} \)
\( = \frac{1}{40} \left[ \ln 1 - \ln \left( \frac{1/4}{9/4} \right) \right] = \frac{1}{40} [-\ln (1/9)] = \frac{1}{40} \ln 9 = \frac{1}{20} \ln 3 \)

Question. Evaluate
(i) \( \int_{a}^{b} \frac{dx}{\sqrt{(x - a)(b - x)}} \)
(ii) \( \int_{a}^{b} \sqrt{(x - a)(b - x)} dx \)
Answer: (i) \( I = \int_{a}^{b} \frac{dx}{\sqrt{(x - a)(b - x)}} \)
Put \( x = a \cos^2 \theta + b \sin^2 \theta \)
\( dx = (b - a) \sin 2\theta d\theta \)
Lower limit: \( a = a \cos^2 \theta + b \sin^2 \theta \Rightarrow \theta = 0 \)
Upper Limit: \( b = a \cos^2 \theta + b \sin^2 \theta \Rightarrow \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \)
\( I = \int_{0}^{\pi/2} \frac{2(b - a) \sin \theta \cos \theta}{(b - a) \sin \theta \cos \theta} d\theta = \pi \)
 (ii) \( I = \int_{a}^{b} \sqrt{(x - a)(b - x)} dx \)
Put \( x = a \cos^2 \theta + b \sin^2 \theta \)
\( dx = (b - a) \sin 2\theta d\theta \)
\( = 2(b - a)^2 \int_{0}^{\pi/2} \sin^2 \theta \cos^2 \theta d\theta \)
\( = \frac{(b - a)^2}{2} \int_{0}^{\pi/2} \sin^2 2\theta d\theta \)
\( = \frac{(b - a)^2}{2} \int_{0}^{\pi/2} \left( \frac{1 - \cos 4\theta}{2} \right) d\theta = \frac{\pi(b - a)^2}{8} \)

Question. Evaluate
(i) \( \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \)
(ii) \( \int_{0}^{\pi/2} \frac{e^{\sin x}}{e^{\sin x} + e^{\cos x}} dx \)
(iii) \( \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx \)
(iv) \( \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin x + \cos x} dx \)
Answer: (i) \( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \) By king property
\( I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \)
\( 2I = \int_{0}^{\pi/2} 1 \cdot dx \Rightarrow 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \)
 (ii) \( I = \int_{0}^{\pi/2} \frac{e^{\sin x}}{e^{\sin x} + e^{\cos x}} dx \)
Use king property
\( I = \int_{0}^{\pi/2} \frac{e^{\cos x}}{e^{\sin x} + e^{\cos x}} dx \)
Add
\( 2I = \int_{0}^{\pi/2} dx \Rightarrow I = \frac{\pi}{4} \)
 (iii) \( I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx \)
King property
\( I = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} dx \)
\( 2I = \int_{0}^{a} 1 \cdot dx \Rightarrow 2I = a \Rightarrow I = \frac{a}{2} \)
 (iv) \( I = \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin x + \cos x} dx \)
King property
\( I = \int_{0}^{\pi/2} \frac{a \cos x + b \sin x}{\sin x + \cos x} dx \)
Add
\( 2I = \int_{0}^{\pi/2} \frac{a \sin x + b \sin x + a \cos x + b \sin x}{\sin x + \cos x} dx \)
\( = \int_{0}^{\pi/2} \frac{(a + b)(\sin x + \cos x)}{\sin x + \cos x} dx \)
\( 2I = (a + b) \frac{\pi}{2} \Rightarrow I = (a + b) \frac{\pi}{4} \)

Question. Evaluate
(i) \( \int_{-1}^{2} \{2x\} dx \) (where {\(*\)} denotes fractional part function)
(ii) \( \int_{0}^{10\pi} (|\sin x| + |\cos x|) dx \)
Answer: (i) \( I = \int_{-1}^{2} \{2x\} dx \)
\( 2x = t \Rightarrow dx = \frac{dt}{2} \Rightarrow I = \frac{1}{2} \int_{-2}^{4} \{t\} dt \)
\( = \frac{1}{2} \int_{-2}^{0} \{t\} dt + \frac{1}{2} \int_{0}^{4} \{t\} dt \)
\( = \frac{1}{2} \int_{-2}^{0} (t - [t]) dt + 2 \int_{0}^{1} t dt \)
\( = \frac{1}{2} \left[ \frac{t^2}{2} \right]_{-2}^{0} - \frac{1}{2} \int_{-2}^{-1} (-2) dt - \frac{1}{2} \int_{-1}^{0} (-1) dt + 1 \)
\( = \frac{1}{4} [0 - 4] + (-1 + 2) + \frac{1}{2} (OH) + 1 = -1 + 1 + \frac{1}{2} + 1 = \frac{3}{2} \)
 (ii) \( I = \int_{0}^{10\pi} (|\sin x| + |\cos x|) dx \)
\( = \int_{0}^{10\pi} |\sin x| dx + \int_{0}^{10\pi} |\cos x| dx \)
\( = 10 \int_{0}^{\pi} |\sin x| dx + 10 \int_{0}^{\pi} |\cos x| dx = 10 \times 2 + 10 \times 2 = 40 \)

Question. If \( f(x) \) is an odd function defined on \( \left[-\frac{T}{2}, \frac{T}{2}\right] \) and has period T, then prove that \( \phi(x) = \int_{0}^{x} f(t) dt \) is also periodic with period T.
Answer: \( F(-x) = -f(x) \)
\( f(x + t) = f(x) \)
\( \phi(x) = \int_{0}^{x} f(t) dt \)
\( \phi(x + T) = \int_{0}^{x+T} f(t) dt = \int_{0}^{x} f(t) dt + \int_{x}^{x+T} f(t) dt \)
\( = \phi(x) + \int_{x}^{T/2} f(t) dt + \int_{T/2}^{2+T} f(t) dt \)
Sub \( u + T = t \Rightarrow du = dt \)
\( = \phi(x) + \int_{x}^{T/2} f(t) dt + \int_{-T/2}^{x} f(u + T) du \)
\( = \phi(x) + \int_{x}^{T/2} f(t) dt + \int_{-T/2}^{x} f(u) du \)
\( = \phi(x) + \int_{x}^{T/2} f(t) dt + \int_{-T/2}^{x} f(t) dt \)
\( = \phi(x) + \int_{-T/2}^{T/2} f(t) dt \to a \text{ as on odd function} \)
\( \phi(x + T) = \phi(x) \)

Question. If \( f(x) = 5^{g(x)} \) and \( g(x) = \int_{2}^{x^2} \frac{t}{\ln(1 + t^2)} dt \) then find the value of \( f'(\sqrt{2}) \)
Answer: \( f(x) = 5^{g(x)} \)
\( f'(x) = 5^{g(x)} \ln 5 \cdot g'(x) \)
\( g(x) = \int_{2}^{x^2} \frac{t}{\ln(1 + t^2)} dt \Rightarrow g'(x) = \frac{x^2}{\ln(1 + x^4)} \cdot 2x \)
\( f'(\sqrt{2}) = 5^{g(\sqrt{2})} \ln 5 \cdot g'(\sqrt{2}) \Rightarrow g(\sqrt{2}) = 0 \)
\( g'(\sqrt{2}) = \frac{2 \cdot 2\sqrt{2}}{\ln 5} = \frac{4\sqrt{2}}{\ln 5} \)
\( f'(\sqrt{2}) = 1 \cdot \ln 5 \cdot \frac{4\sqrt{2}}{\ln 5} = 4\sqrt{2} \)

Question. If \( f(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} dt \) then prove that \( f'(x) = 0 \quad \forall x \in \mathbb{R} \).
Answer: \( f(x) = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} dt \)
\( f'(x) = \sin^{-1}(\sin x) (2 \sin x \cos x) + \cos^{-1}(\cos x) \cdot (-2 \cos x \sin x) \)
\( = x (\sin 2x) + x (-\sin 2x) \)
\( f'(x) = 0 \)

Question. Evaluate
(i) \( \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2 - r^2}} \)
(ii) \( \lim_{n \to \infty} \frac{3}{n} \left[ 1 + \sqrt{\frac{n}{n+3}} + \sqrt{\frac{n}{n+6}} + \sqrt{\frac{n}{n+9}} + \dots + \sqrt{\frac{n}{n+3(n-1)}} \right] \)
Answer: (i) \( I = \lim_{h \to \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2 - r^2}} \)
\( = \lim_{h \to \infty} \frac{1}{n} \sum_{r=1}^{n-1} \frac{1}{\sqrt{1 - \left(\frac{r}{n}\right)^2}} \)
\( = \int_{0}^{1} \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x \Big|_{0}^{1} \)
\( = \sin^{-1} 1 - 0 = \frac{\pi}{2} \)
(ii) \( \lim_{n \to \infty} \frac{3}{n} \left[ 1 + \sqrt{\frac{n}{n+3}} + \sqrt{\frac{n}{n+6}} + \sqrt{\frac{n}{n+9}} + \dots + \sqrt{\frac{n}{n+3(n-1)}} \right] \)
\( = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n} \sqrt{\frac{n}{n+3r}} \)
\( = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n} \frac{1}{\sqrt{1 + 3\left(\frac{r}{n}\right)}} = 3 \int_{0}^{1} \frac{dx}{\sqrt{1 + 3x}} dx = 2 \)

Question. \( \int_{0}^{\pi} e^{\cos^2 x} \cos^3 (2n+1)x dx \), \( n \in I \)
Answer: \( I = \int_{0}^{\pi} e^{\cos^2 x} \cos^3 (2n+1)x dx \)
If \( n \in \text{even integer assume } n = 2 \)
\( I = \int_{0}^{\pi} \underbrace{e^{\cos^2 x} \cos^3 5x}_{f(x)} dx \)
\( f(\pi - x) = -f(x) \)
\( I = 0 \)
If \( n \in \text{odd integer assume } n = 1 \)
\( I = \int_{0}^{\pi} e^{\cos^2 x} \cos^3 3x dx \)
By applying again given \( f(\pi - n) = -f(n) \)
\( I = 0 \)

Question. If \( f, g, h \) be continuous function on \( [0, a] \) such that \( f(a - x) = f(x) \), \( g(a - x) = -g(x) \) and \( 3h(x) - 4h(a - x) = 5 \), then prove that, \( \int_{0}^{a} f(x) g(x) h(x) dx = 0 \).
Answer: \( I = \int_{0}^{a} f(x) g(x) h(x) dx \)
\( = \int_{0}^{a} f(a - x)g(a - x)h(a - x) dx \)
\( I = \int_{0}^{a} f(x)[-g(x)] \left[ \frac{3h(x) - 5}{4} \right] dx \)
\( = -\frac{3}{4} \int_{0}^{a} f(x)g(x)h(x) dx + \frac{5}{4} \int_{0}^{a} f(x)g(x) dx \)
By using again queen
\( I = -\frac{3}{4} I \Rightarrow I = 0 \)

Question. Show that \( \int_{0}^{x} e^{zx} \cdot e^{-z^2} dz = e^{x^2/4} \int_{0}^{x} e^{-z^2/4} dz \).
Answer: \( I = \int_{0}^{x} e^{(x - z)z} dz \quad (x \to \text{constant}, z \to \text{Variable}) \)
Put \( z = \frac{x + t}{2} \)
\( = \frac{1}{2} \int_{-x}^{x} e^{\left(x\left(\frac{x+t}{2}\right) - \left(\frac{x+t}{2}\right)^2\right)} dt = \frac{1}{2} \int_{-x}^{x} e^{\left(\frac{x^2 - t^2}{4}\right)} dt \)
\( = \frac{1}{2} e^{x^2/4} \int_{-x}^{x} e^{-t^2/4} dt = e^{x^2/4} \int_{0}^{x} e^{-t^2/4} dt \)

Question. Let \( f(x) = \begin{cases} 1 - x & \text{if } 0 \le x \le 1 \\ 0 & \text{if } 1 < x \le 2 \\ (2 - x)^2 & \text{if } 2 < x \le 3 \end{cases} \). Define the function \( F(x) = \int_{0}^{x} f(t) dt \) and show that F is continuous in \( [0, 3] \) and differentiable in \( (0, 3) \).
Answer: \( F(x) = \begin{cases} \int_{0}^{x} (1 - x) dx = x - \frac{x^2}{2} & ; x \in [0, 1] \\ \int_{0}^{1} (1 - x) dx + \int_{1}^{x} 0 dx = \frac{1}{2} & ; x \in (1, 2] \\ \int_{0}^{1} (1 - x) dx + \int_{1}^{2} 0 dx + \int_{2}^{x} (2 - x)^2 dx = \frac{1}{2} - \frac{1}{3} (2 - x)^3 & ; x \in (2, 3] \end{cases} \)
check the continuity at \( x = 1, 2 \)
\( F(1^-) = \frac{1}{2}, F(1^+) = \frac{1}{2}, F(1) = \frac{1}{2} \)
\( F(2^-) = \frac{1}{2}, F(2^+) = \frac{1}{2}, F(2) = \frac{1}{2} \)
Hence \( F(x) \) is continuous \( [0, 3] \)
check the differentiability at \( x = 1, 2 \)
\( F'(x) = \begin{cases} 1 - x & \text{if } 0 < x < 1 \\ 0 & \text{if } 1 < x < 2 \\ (2 - x)^2 & \text{if } 2 < x < 3 \end{cases} \)
\( F'(1^-) = 0, F'(1^+) = 0, F'(1) = 0 \)
\( F'(2^-) = 0, F'(2^+) = 0, F'(2) = 0 \)
Hence \( F(x) \) is differentiable at \( x \in (0, 3) \)

Question. Evaluate, \( \int_{0}^{1} |x - t| \cdot \cos \pi t dt \) where 'x' is any real number
Answer: There are 3 cases
Case-I when \( 0 \le x \le 1 \Rightarrow \int_{0}^{1} |x - t| \cdot \cos \pi t dt \)
\( = \int_{0}^{x} (x - t) \cos \pi t dt + \int_{x}^{1} (t - x) \cos \pi t dt \)
[Using by parts \( \uparrow \)]
\( = - \frac{2 \cos \pi x}{\pi^2} \)
 Case-II When \( x < 0 \Rightarrow \int_{0}^{1} |x - t| \cdot \cos \pi t dt \)
\( = \int_{0}^{1} (t - x) \cos \pi t dt = - \frac{2}{\pi^2} \)
 Case-III when \( x > 1 \Rightarrow \int_{0}^{1} |x - t| \cdot \cos \pi t dt \)
\( = \int_{0}^{1} (x - t) \cos \pi t dt = \frac{2}{\pi^2} \)

Question. Evaluate, \( I = \int_{0}^{1} 2 \sin (p t) \sin (q t) dt \), if :
(i) p & q are different roots of the equation, \( \tan x = x \).
(ii) p & q are equal and either is root of the equation \( \tan x = x \).
Answer: (a) If P & q are diff. roots of the equation \( \tan P = P \), \( \tan q = q \)
\( I = \int_{0}^{1} 2 \sin pt \sin(qt) dt \)
Integrates by using by parts taking \( \sin pt \) as second function.
\( = - \left[ 2 \sin qt \frac{\cos pt}{p} \right]_{0}^{1} + \int_{0}^{1} 2q \cos t \left( \frac{\cos pt}{p} \right) dt \)
\( = - \frac{2}{p} \sin q \cos p + \frac{2q}{p} \int_{0}^{1} \cos t \cos pt dt \)
\( = - \frac{2}{p} \sin q \cos p + \frac{2q}{p} \left[ \cos t \frac{\sin pt}{p} \Big|_{0}^{1} + \frac{q}{p} \int_{0}^{1} \sin p \sin t dt \right] \)
\( = - \frac{2}{p} \sin q \cos p + \frac{2q}{p} \sin po \cos q + \frac{q^2}{p^2} I \)
\( I \left( 1 - \frac{q^2}{p^2} \right) = \frac{2q}{p} \sin \cos q - \frac{2}{p} \sin q \cos p \)
 (b) Given \( p = q \)
\( I = \int_{0}^{1} 2 \sin^2 p \cdot t dt \)
\( = \int_{0}^{1} (1 - \cos 2pt) dt = \left[ t - \frac{\sin 2pt}{2p} \right]_{0}^{1} \)
\( = 1 - \frac{\sin 2p}{2p} = 1 - \frac{(2\tan p)/(1 + \tan^2 p)}{2p} \)
\( = 1 - \frac{2p}{(1 + p^2)2p} = \frac{p^2}{1 + p^2} \)

Advanced Subjective Questions

Question. \( \int_{0}^{2\pi} e^{x} \cos \left( \frac{\pi}{4} + \frac{x}{2} \right) dx \)
Answer: \[ \int_{0}^{2\pi} e^{x} \left( \frac{1}{\sqrt{2}} \cos \frac{x}{2} - \frac{1}{\sqrt{2}} \sin \frac{x}{2} \right) dx \] \[ = \frac{1}{\sqrt{2}} \int_{0}^{2\pi} e^{x} \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) dx \]

Question. \( \int_{0}^{\pi/4} \frac{\cos x - \sin x}{10 + \sin 2x} dx \)
Answer: \( I = \int_{0}^{\pi/4} \frac{\cos x - \sin x}{10 + \sin 2x} dx \) \( = \int_{0}^{\pi/4} \frac{\cos x - \sin x}{9 + (\cos x + \sin x)^{2}} dx \) Let \( \sin x + \cos x = t \) \( (\cos x - \sin x) dx = dt \)

Question. \( \int_{0}^{\pi} \frac{(ax + b)\sec x \tan x}{4 + \tan^{2} x} dx \), (a, b > 0)
Answer: Use king and add. \( 2I = \int_{0}^{\pi} (a\pi + 2b) \frac{\sec x \tan x}{3 + \sec^{2} x} dx \) Let \( \sec x = t \) \( (\sec x \cdot \tan x) \cdot dx = dt \)

Question. \( \int_{0}^{\pi} \frac{(2x + 3)\sin x}{(1 + \cos^{2} x)} dx \)
Answer: Use king and add. \( 2I = (2\pi + 3) \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx \) Let \( \cos x = t \) \( (\sin x) \cdot dx = -dt \)

Question. Show that \( \int_{0}^{p+q\pi} |\cos x| dx = 2q + \sin p \) where \( q \in \mathbb{N} \) & \( -\frac{\pi}{2} < p < \frac{\pi}{2} \)
Answer: Let \( I = \int_{0}^{p+q\pi} |\cos x| dx \) \( = \int_{0}^{q\pi} |\cos x| dx + \int_{q\pi}^{p+q\pi} |\cos x| dx \) \( = q \int_{0}^{\pi} |\cos x| dx + \int_{0}^{p} |\cos x| dx \) { \( \because \) period of \( |\cos x| \) is \( \pi \) } \( = q \left\{ \int_{0}^{\pi/2} |\cos x| dx + \int_{\pi/2}^{\pi} |\cos x| dx \right\} + \int_{0}^{p} \cos x dx \) \( = q \left\{ \int_{0}^{\pi/2} \cos x dx - \int_{\pi/2}^{\pi} \cos x dx \right\} + \int_{0}^{p} \cos x dx \) \( = q \{ (\sin x)_{0}^{\pi/2} - (\sin x)_{\pi/2}^{\pi} \} + (\sin x)_{0}^{p} \) \( = q \{ (1 - 0) - (0 - 1) \} + \sin p - \sin 0 = 2q + \sin p \)

Question. \( \int_{0}^{1} \frac{\sin^{-1} \sqrt{x}}{x^{2} - x + 1} dx \)
Answer: \( I = \int_{0}^{1} \frac{\sin^{-1} \sqrt{1 - x}}{x^{2} - x + 1} dx \) Let \( \sqrt{x} = w \) \( I = \int_{0}^{1} \frac{2w}{w^{4} - w^{2} + 1} dw \) Let \( w^{2} = z \) \( I = \int_{0}^{1} \frac{dz}{z^{2} - z + 1} \)

Question. \( \int_{0}^{\pi/2} \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right] dx \)
Answer: \( I = \int_{0}^{\pi/2} \tan^{-1} \left[ \frac{|\sin x/2 + \cos x/2| + |\sin x/2 - \cos x/2|}{|\sin x/2 + \cos x/2| - |\sin x/2 - \cos x/2|} \right] dx \) \( = \int_{0}^{\pi/4} \tan^{-1} (\cot x/2) dx + \int_{\pi/4}^{\pi/2} \tan^{-1} (\cot x/2) dx \)

Question. \( \int_{\sqrt{\frac{3a^{2} + b^{2}}{2}}}^{\sqrt{\frac{a^{2} + b^{2}}{2}}} \frac{x \cdot dx}{(x^{2} - a^{2})(b^{2} - x^{2})} \)
Answer: Let \( x^{2} = t \) \( I = \frac{1}{2} \int_{\frac{3a^{2} + b^{2}}{2}}^{\frac{a^{2} + b^{2}}{2}} \frac{dt}{(t - a^{2})(b^{2} - t)} \) \( = \frac{1}{2(b^{2} - a^{2})} \int_{\frac{3a^{2} + b^{2}}{2}}^{\frac{a^{2} + b^{2}}{2}} \left\{ \frac{1}{(t - a^{2})} + \frac{1}{(b^{2} - t)} \right\} dt \)

Question. Comment upon the nature of roots of the quadratic equation \( x^{2} + 2x = k + \int_{0}^{1} |t + k| dt \) depending on the value of \( k \in \mathbb{R} \).
Answer: \( x^{2} + 2x = k + \int_{0}^{-k} -(t + k) dt + \int_{-k}^{1} (t + k) dt \) \( = 2k + k^{2} + 1/2 \) \( \implies 2x^{2} + 4x - (2k^{2} + 4k + 1) = 0 \) \( D = 16 + 4(2k^{2} + 4k + 1) \cdot 2 \) \( D' < 0 \) so \( D > 0 \implies \) Real & Distinct roots \( \forall x \in \mathbb{R} \).

Question. \( \int_{0}^{2a} x \sin^{-1} \left( \frac{1}{2} \sqrt{\frac{2a - x}{a}} \right) dx \)
Answer: Let \( \sin^{-1} \left( \frac{1}{2} \sqrt{\frac{2a - x}{a}} \right) = t \implies x = 2a - 4a \sin^{2} t \) \( I = \int_{\pi/4}^{0} (2a - 4a \sin^{2} t) \cdot t \cdot (-8a \sin t \cos t) dt \) \( = a^{2} \int_{0}^{\pi/4} (16 \sin t \cos t - 32 \sin^{3} t \cos t) \cdot t \cdot dt \) using king & add & then \( (\pi/4 - t) = w \) \( 2I = 16a^{2} \int_{0}^{\pi/4} (\sin w \cos w - 2 \sin^{3} w \cos w) \cdot \left( \frac{\pi}{4} \right) (-dw) \) \( I = 2a^{2} \int_{0}^{\pi/4} (\sin w \cos w - 2 \sin^{3} w \cos w) dw \)

Question. Let \( u = \int_{0}^{\pi/4} \left( \frac{\cos x}{\sin x + \cos x} \right)^{2} dx \) and \( v = \int_{0}^{\pi/4} \left( \frac{\sin x + \cos x}{\cos x} \right)^{2} dx \). Find the value of \( \frac{v}{u} \).
Answer: \( u = \int_{0}^{\pi/4} \frac{1}{(1 + \tan x)^{2}} \cdot \frac{\sec^{2} x}{(\sec^{2} x)} dx \) \( u = \int_{0}^{1} \frac{dt}{(1 + t^{2})(1 + t)^{2}} = \frac{1}{2} \int_{0}^{1} \left( \frac{1}{1 + t^{2}} + \frac{1}{(1 + t)^{2}} \right) dt \) ...(i) \( v = \int_{0}^{\pi/4} (1 + \tan x)^{2} dx = \int_{0}^{\pi/4} (2 + \sec^{2} x + 2 \tan x) dx \)

Question. \( \int_{0}^{2\pi} \frac{x^{2} \sin x}{8 + \sin^{2} x} dx \)
Answer: \( I = \int_{0}^{2\pi} x^{2} \left( \frac{\sin x}{9 - \cos^{2} x} \right) dx \) (using by parts) where \( x^{2} \) is I and the other term is II. IInd let \( \cos x = t \)

Question. \( \int_{0}^{\pi/4} \frac{x^{2} (\sin 2x - \cos 2x)}{(1 + \sin 2x) \cos^{2} x} dx \)
Answer: \( = \int_{0}^{\pi/4} \frac{x^{2} ((1 + \sin 2x) - (1 + \cos 2x))}{(1 + \sin 2x) \cos^{2} x} dx \) \( = \int_{0}^{\pi/4} x^{2} (\sec^{2} x - \sec^{2}(\pi/4 - x)) dx \) Using by parts take \( x^{2} \) as first function \( = \left. x^{2}(\tan x + \tan(\pi/4 - x)) \right|_{0}^{\pi/4} - 2 \int_{0}^{\pi/4} x(\tan x + \tan(\pi/4 - x)) dx \) \( = \frac{\pi^{2}}{16} - 2 \int_{0}^{\pi/4} x dx + 2 \int_{0}^{\pi/4} x \tan x \tan \left( \frac{\pi}{4} - x \right) dx \) \( \left\{ \because \frac{\pi}{4} = x + \left( \frac{\pi}{4} - x \right) \right\} \) \( = \frac{\pi^{2}}{16} - \frac{\pi^{2}}{16} + 2 \int_{0}^{\pi/4} x \tan x \tan(\pi/4 - x) dx \) use king & add \( I = \frac{\pi}{4} \int_{0}^{\pi/4} \frac{\tan x (1 - \tan x)}{(1 + \tan x)} dx \) \( = \frac{\pi}{4} \int_{0}^{\pi/4} \left( -\tan x + 2 - \frac{2}{1 + \tan x} \right) dx \) \( = \frac{\pi}{4} (-\ln \sec x) - \frac{\pi}{4} (x)_{0}^{\pi/4} - \frac{\pi}{2} \int_{0}^{\pi/4} \frac{dx}{1 + \tan x} \) \( = \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 \)

Question. Prove that \( \int_{0}^{x} \left( \int_{0}^{u} f(t) dt \right) du = \int_{0}^{x} f(u) \cdot (x - u) du \)
Answer: Using by parts \( \int_{0}^{x} \left( 1 \int_{0}^{u} f(t) dt \right) du = \left[ \int_{0}^{u} f(t) dt \cdot u \right]_{0}^{x} - \int_{0}^{x} \{f(u) \cdot u\} du \) \( = x \int_{0}^{x} f(t) dt - \int_{0}^{x} u f(u) du \) \( = x \int_{0}^{x} f(u) du - \int_{0}^{x} u f(u) du \) \( = \int_{0}^{x} (x - u) f(u) du \)

Question. \( \int_{0}^{\pi} \frac{dx}{(5 + 4 \cos x)^{2}} \)
Answer: \( \frac{1}{4 \sin x} \frac{d}{dx} \left( \frac{1}{5 + 4 \cos x} \right) = \frac{1}{(5 + 4 \cos x)^{2}} \) so \( I = \int \frac{1}{4 \sin x} \cdot \frac{d}{dx} \left( \frac{1}{5 + 4 \cos x} \right) dx \) (By parts).

Question. Evaluate \( \int_{0}^{1} \ln(\sqrt{1 - x} + \sqrt{1 + x}) dx \)
Answer: \( I = \int_{0}^{1} \ln(\sqrt{1 - x} + \sqrt{1 + x}) \cdot dx \times \frac{2}{2} \) \( I = \frac{1}{2} \int_{0}^{1} \ln(2 + 2\sqrt{1 - x^{2}}) dx \) \( = \frac{1}{2} \int_{0}^{1} \ln 2 dx + \int_{0}^{1} \ln(1 + \sqrt{1 - x^{2}}) dx \) using by parts & take 1 as IInd function so \( I_{1} = \int_{0}^{1} 1 \cdot \ln(1 + \sqrt{1 - x^{2}}) dx \) \( = \left. x \ln(1 + \sqrt{1 - x^{2}}) \right|_{0}^{1} - \int_{0}^{1} \frac{(-x^{2})}{\sqrt{1 - x^{2}} (1 + \sqrt{1 - x^{2}})} dx \) Let \( x = \sin \theta \) & so on

Question. \( \int_{1}^{16} \tan^{-1} \sqrt{\sqrt{x} - 1} dx \)
Answer: put \( \sqrt{\sqrt{x} - 1} = t \implies \sqrt{x} - 1 = t^{2} \implies x = (t^{2} + 1)^{2} \) \( dx = 2(t^{2} + 1) 2t dt \) \( I = \int_{0}^{\sqrt{3}} 4t(t^{2} + 1) \tan^{-1} t dt \) \( = \int_{0}^{\sqrt{3}} (4t^{3} + 4t) \tan^{-1} t dt \) use by parts \( = \frac{16\pi}{3} - 2\sqrt{3} \)

Question. \( \lim_{n \to \infty} n^{2} \int_{-1/n}^{1/n} (2006 \sin x + 2007 \cos x) |x| dx \)
Answer: \( I = \lim_{n \to \infty} n^{2} \int_{-1/n}^{1/n} 2006 \sin x |x| dx + \int_{-1/n}^{1/n} 2007 \cos x |x| dx \) The first integral is odd, the second integral is even. so \( I = \lim_{n \to \infty} 2 \int_{0}^{1/n} n^{2} \cdot 2007 \cos x \cdot x dx \) \( \lim_{n \to \infty} 4014 \left( \frac{\sin 1/n}{1/n} + \frac{\cos 1/n - 1}{1/n^{2}} \right) = 2007 \)

Question. Show that \( \int_{0}^{\infty} f \left( \frac{a}{x} + \frac{x}{a} \right) \frac{\ln x}{x} dx = \ln a \cdot \int_{0}^{\infty} f \left( \frac{a}{x} + \frac{x}{a} \right) \frac{dx}{x} \)
Answer: Let \( a/x = t \) \( = \int_{0}^{\infty} f \left( t + \frac{1}{t} \right) \frac{(\ln a - \ln t)}{t} dt \) \( = \ln a \int_{0}^{\infty} f \left( t + \frac{1}{t} \right) \frac{dt}{t} - \int_{0}^{\infty} f \left( t + \frac{1}{t} \right) \frac{\ln t}{t} dt \) For the second integral, let \( t = a/x \implies 0 \) \( = \ln a \int_{0}^{\infty} f \left( \frac{a}{x} + \frac{x}{a} \right) \frac{dx}{x} \)

Question. Evaluate the definite integral, \( \int_{-1}^{1} \frac{(2x^{332} + x^{998} + 4x^{1668} \sin x^{691})}{1 + x^{666}} dx \).
Answer: \( I = \int_{-1}^{1} \left( \frac{2x^{332} + x^{998}}{1 + x^{666}} + \frac{4x^{1668} \sin x^{691}}{1 + x^{666}} \right) dx \) First term is even, second term is odd. so \( I = 2 \int_{0}^{1} \frac{x^{332}}{1 + x^{666}} dx + \int_{0}^{1} \frac{x^{332} (1 + x^{666})}{1 + x^{666}} dx \) \( x^{333} = t \) & so on.

Question. For \( a \ge 2 \), if the value of the definite integral \( \int_{0}^{\infty} \frac{dx}{a^{2} + \left( x - \frac{1}{x} \right)^{2}} \) equals \( \frac{\pi}{5050} \). Find the value of a.
Answer: \( \int_{0}^{\infty} \frac{dx}{a^{2} + (x - 1/x)^{2}} \) \( = \frac{1}{2} \int_{0}^{\infty} \frac{(1 + 1/x^{2}) + (1 - 1/x^{2})}{a^{2} + (x - 1/x)^{2}} dx \) \( = \frac{1}{2} \left\{ \int_{0}^{\infty} \frac{(1 + 1/x^{2}) dx}{(a^{2}) + (x - 1/x)^{2}} + \int_{0}^{\infty} \frac{(1 - 1/x^{2}) dx}{(a^{2} - 4) + (x + 1/x)^{2}} \right\} \) let \( x - 1/x = t \) for the first, let \( x + 1/x = t \) for the second.