Definite Integration JEE Mathematics Worksheets Set 02

Subjective Questions

Question. If \( f(x) = \frac{\sin x}{x} \quad \forall x \in (0, \pi] \), prove that,
\( \frac{\pi}{2} \int_{0}^{\pi/2} f(x) f\left(\frac{\pi}{2} - x\right) dx = \int_{0}^{\pi} f(x) dx \)
Answer: \( f\left(\frac{\pi}{2} - x\right) = \frac{\cos x}{\left(\frac{\pi}{2} - x\right)} \)
\( \frac{\pi}{2} \int_{0}^{\pi/2} f(x) f\left(\frac{\pi}{2} - x\right) dx = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\sin x}{x} \frac{\cos x}{\left(\frac{\pi}{2} - x\right)} dx \)
\( = \pi \int_{0}^{\pi/2} \frac{\sin 2x}{2x(x - 2\pi)} dx \)
\( = \pi \left[ \int_{0}^{\pi/2} \frac{\sin 2x}{2x} dx + \int_{0}^{\pi/2} \frac{\sin 2x}{x - 2x} dx \right] \)
\( = 2 \int_{0}^{\pi/2} \frac{\sin 2x}{2x} dx \)
Put \( 2x = t \Rightarrow 2dx = dt \)
\( = \int_{0}^{\pi} \frac{\sin t}{t} dt = \int_{0}^{\pi} \frac{\sin x}{x} dx = \int_{0}^{\pi} f(x) dx \)

Question. Evaluate
\( \int_{0}^{1} \frac{dx}{(5 + 2x - 2x^2)(1 + e^{2-4x})} \)
Answer: \( I = \int_{0}^{1} \frac{dx}{(5 + 2x - 2x^2)(1 + e^{2-4x})} \)
Use king's property
\( I = \int_{0}^{1} \frac{dx}{[5 + 2(1 - x) - 2(1 - x)^2](1 + e^{2 - 4(1 - x)})} \)
\( I = \int_{0}^{1} \frac{dx}{(5 + 2x - 2x^2)(1 + e^{-(2 - 4x)})} \)
\( 2I = \int_{0}^{1} \frac{e^{2 - 4x} + 1}{(5 + 2x - 2x^2)(1 + e^{2 - 1x})} dx \)
\( 2I = \int_{0}^{1} \frac{dx}{5 + 2x - 2x^2} = \frac{1}{4} \int_{0}^{1} \frac{dx}{\frac{5}{2} + x - x^2} \)
\( = \frac{1}{4} \int_{0}^{1} \frac{dx}{\frac{11}{4} - \left(x - \frac{1}{2}\right)^2} \)
\( = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11} + 1}{\sqrt{11} - 1} \right) \)

Question. If \( n > 1 \), evaluate \( \int_{0}^{\infty} \frac{dx}{(x + \sqrt{1 + x^2})^n} \)
Answer: \( I = \int_{0}^{\infty} \frac{dx}{(x + \_\sqrt{1 + x^2})^n} \)
Put \( x = \tan\theta \)
\( dx = \sec^2\theta d\theta \)
\( = \int_{0}^{\pi/2} \frac{\sec^2\theta d\theta}{(\tan\theta + \sec\theta)^n} = \int_{0}^{\pi/2} \frac{\cos^{n-2}\theta d\theta}{(1 + \sin\theta)^n} \)
King
\( I = \int_{0}^{\pi/2} \frac{\sin^{n-2}\theta d\theta}{(1 + \cos\theta)^n} = \int_{0}^{\pi/2} \frac{\left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)^{n-2} d\theta}{\left(2\cos^2\frac{\theta}{2}\right)^n} \)
\( = \frac{1}{4} \int_{0}^{\pi/2} \frac{\sin^{n-2}(\theta/2)}{\cos^{n+2}(\theta/2)} d\theta \)
\( = \frac{1}{4} \int_{0}^{\pi/2} \frac{\sin^{n-2}(\theta/2)}{\cos^{n-2}(\theta/2)} \cdot \frac{1}{\cos^4(\theta/2)} d\theta \)
\( I = \frac{1}{4} \int_{0}^{\pi/2} \tan^{n-2}\left(\frac{\theta}{2}\right) \left(1 + \tan^2\left(\frac{\theta}{2}\right)\right) \sec^2\left(\frac{\theta}{2}\right) d\theta \)
put \( \tan\frac{\theta}{2} = t \)
\( \frac{1}{2}\sec^2\frac{\theta}{2} d\theta = dt \)
\( = \frac{1}{2} \int_{0}^{1} t^{n-2}(1 + t^2)dt = \frac{1}{2} \int_{0}^{1} (t^{n-2} + t^n)dt \)
\( = \frac{1}{2} \left[ \frac{t^{n-1}}{n-1} + \frac{t^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{2} \left[ \frac{1}{n-1} + \frac{1}{n+1} \right] = \frac{n}{n^2 - 1} \)

Question. \( \int_{0}^{1} (\{2x\} - 1) (\{3x\} - 1) dx \),
where {\(*\)} denotes fractional part of x.
Answer: \( I = \int_{0}^{1} (\{2x\} - 1) (\{3x\} - 1) dx \)
\( = \int_{0}^{1} (2x - [2x] - 1) (3x - [3x] - 1) dx \)
\( = \int_{0}^{1/3} (2x - 1) (3x - 1) dx + \int_{1/3}^{1/2} (2x - 1) (3x - 2) dx + \int_{1/2}^{2/3} (2x - 2) (3x - 2) dx + \int_{2/3}^{1} (2x - 2) (3x - 3) dx \)
\( = \frac{19}{72} \)

Question. Let \( f(x) \) be a continuous function \( \forall x \in \mathbb{R} \), except at \( x = 0 \) such that \( \int_{0}^{a} f(x) dx \), \( a \in \mathbb{R}^+ \) exists.
If \( g(x) = \int_{x}^{a} \frac{f(t)}{t} dt \), prove that \( \int_{0}^{a} g(x) dx = \int_{0}^{a} f(x) dx \).
Answer: \( g(x) = \int_{x}^{a} \frac{f(t)}{t} dt \)
\( g'(x) = -\frac{f(x)}{x} \)
\( x g'(x) = -f'(x) \)
Integrate both the side w.r.t.x.
\( \int_{0}^{a} x \cdot g'(x)dx = -\int_{0}^{a} f(x)dx \)
\( x g(x) \big|_{0}^{a} - \int_{0}^{a} g(x)dx = -\int_{0}^{a} f(x)dx \)
\( g(x) - \int_{0}^{a} g(x)dx = -\int_{0}^{a} f(x)dx \)
\( g(x) = \int_{0}^{a} \frac{f(t)}{t} dt = 0 \)
\( g(a) = \int_{0}^{a} \frac{f(t)}{t} dt = 0 \)
\( -\int_{0}^{a} g(x)dx = -\int_{0}^{a} f(x)dx \)
\( \int_{0}^{a} g(x)dx = \int_{0}^{a} f(x)dx \)

Question. \( \int_{0}^{\pi} \frac{x dx}{9\cos^2 x + \sin^2 x} \)
Answer: \( I = \int_{0}^{\pi} \frac{x dx}{9\cos^2 x + \sin^2 x} \)
\( I = \int_{0}^{\pi} \frac{(\pi - x)dx}{9\cos^2 x + \sin^2 x} \) king
add
\( 2I = \pi \int_{0}^{\pi} \frac{dx}{9\cos^2 x + \sin^2 x} \)
queen
\( 2I = 2\pi \int_{0}^{\pi/2} \frac{dx}{9\cos^2 x + \sin^2 x} \)
\( I = \pi \int_{0}^{\pi/2} \frac{dx}{9\cos^2 x + \sin^2 x} \)
\( I = \pi \int_{0}^{\pi/2} \frac{\sec^2 x dx}{\tan^2 x + 9} \)
Put \( \tan x = t \Rightarrow \sec^2 x dx = dt \)
\( = \pi \int_{0}^{\infty} \frac{dt}{a + t^2} = \frac{\pi}{3} \tan^{-1} \frac{t}{3} \Big|_{0}^{\infty} = \frac{\pi}{3} \left[ \frac{\pi}{2} \right] = \frac{\pi^2}{6} \)

Question. \( \int_{0}^{\pi/2} \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx \)
Answer: \( I = \int_{0}^{\pi/2} \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx \)
\( = \int_{0}^{\pi/2} \sqrt{\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^2} dx = \int_{0}^{\pi/2} \left| \frac{\cos x - \sin x}{\cos x + \sin x} \right| dx \)
By using queen properly
\( I = 2 \int_{0}^{\pi/2} \left| \frac{\cos x - \sin x}{\cos x + \sin x} \right| dx \)
Let \( \cos x + \sin x = t \)
\( (\cos x - \sin x)dx = dt \)
\( = 2 \int_{1}^{\sqrt{2}} \frac{dt}{t} = 2[\ln t]_{1}^{\sqrt{2}} \)
\( I = 2[\ln\sqrt{2}] \Rightarrow I = \ln 2 \)

Question. Evaluate \( I_n = \int_{1}^{e} (\ln x)^n dx \) hence find \( I_3 \).
Answer: \( I_n = \int_{1}^{e} ((\ln x)^n) dx \)
\( = x(\ln x)^n \big|_{1}^{e} - \int_{1}^{e} n(\ln x)^{n-1} \cdot \frac{1}{x} \cdot x dx \)
\( I_n = e(\ln e)^n - n I_{n-1} \)
\( I_n + n I_{n-1} = e \)
Put \( n = 1, 2, 3 \) respectively
\( I_3 + 3I_2 = e \quad \dots(1) \)
\( I_2 + 2I_1 = e \quad \dots(2) \)
\( I_1 + I_0 = e \quad \dots(3) \)
\( I_0 = e - 1 \)
\( I_1 = 1 \)
\( I_2 = 0 - 2 \)
\( I_3 = 6 - 2e \)

Question. \( \int_{0}^{\pi/2} \sin 2x \cdot \text{arc}\tan(\sin x) dx \)
Answer: \( I = \int_{0}^{\pi/2} 2\sin x \cos x \tan^{-1}(\sec x) dx \)
Put \( \sin x = t \)
\( \cos x dx = dt \)
\( = 2 \int_{0}^{1} t(\tan^{-1} x) dt = 2 \left[ \frac{t^2}{2} \tan^{-1} t \Big|_{0}^{1} - \int \frac{t^2}{1 + t^2} dt \right] \)
\( = t^2 \tan^{-1} t - \int 1.dt \int \frac{dt}{1 + t^2} \)
\( = t^2 \tan^{-1} t - t + \tan^{-1} t \big|_{0}^{1} = \frac{\pi}{2} - 1 \)

Question. \( \int_{0}^{\pi/4} \frac{x dx}{\cos x(\cos x + \sin x)} \)
Answer: \( I = \int_{0}^{\pi/4} \frac{x dx}{\cos x(\cos x + \sin x)} \)
\( = \int_{0}^{\pi/4} \frac{x dx}{\cos^2 x + \cos x \sin x} = \int_{0}^{\pi/4} \frac{x dx}{\frac{1 + \cos 2x}{2} + \frac{\sin 2x}{2}} \)
\( I = 2 \int_{0}^{\pi/4} \frac{x dx}{1 + \cos 2x + \sin 2x} \)
King's
\( I = 2 \int_{0}^{\pi/4} \left( \frac{\frac{3}{4} - x}{1 + \sin 2x + \cos 2x} \right) dx \)
\( 2I = 2 \times \frac{\pi}{4} \int_{0}^{\pi/4} \frac{dx}{1 + \sin 2x + \cos 2x} \)
\( I = \frac{\pi}{4} \int_{0}^{\pi/4} \frac{dx}{2\cos^2 x + 2\sin x \cos x} \)
\( = \frac{\pi}{4} \int_{0}^{\pi/4} \frac{dx}{2\cos^2 x(1 + \tan x)} dx \)
put \( 1 + \tan x = t \)
\( \sec^2 x dx = dt \)
\( = \frac{\pi}{8} \ln(1 + \tan x) \big|_{0}^{\pi/4} \)
\( I = \frac{\pi}{8} \ln 2 \)

Question. \( \int_{1}^{2} \frac{(x^2 - 1)dx}{x^3 \cdot \sqrt{2x^4 - 2x^2 + 1}} = \frac{u}{v} \) where u and v are in their lowest form. Find the value of \( \frac{(1000)u}{v} \).
Answer: \( I = \int_{1}^{2} \frac{(x^2 - 1)dx}{x^3 \sqrt{(x^2)^2 + (x^2 - 1)^2}} \)
\( = \int_{1}^{2} \frac{dx}{x^3 \sqrt{ \left(\frac{x^2}{x^2 - 1}\right)^2 + 1 }} \)
Let \( \frac{x^2}{x^2 - 1} = t \)
\( \frac{x^2 - 1}{x^2} = \frac{1}{t} \)
\( 1 - \frac{1}{x^2} = \frac{1}{t} \)
\( \frac{2}{x^3} dx = -\frac{1}{t^2} dt \)
\( \pm - \int_{\infty}^{4/3} \frac{dt}{2t^2 \sqrt{t^2 + 1}} = \int_{\infty}^{4/3} \frac{dt}{2t^3 \sqrt{1 + \frac{1}{t^2}}} \)
Let \( 1 + \frac{1}{t^2} = u \Rightarrow -\frac{2}{t^3} dt = du \)
\( = \frac{1}{4} \int_{1}^{25/16} \frac{du}{\sqrt{u}} = \frac{1}{4} [2\sqrt{u}]_{1}^{25/16} \)
\( = \frac{1}{2} \left[ \frac{25}{16} - 1 \right] = \frac{1}{2}\left(\frac{5}{4} - 1\right) = \frac{1}{8} \Rightarrow \frac{u}{v} = \frac{1}{8} \)
\( (1000) \times \frac{u}{v} = \frac{1000}{8} = 125 \)

Question. Find the value of the definite integral
\( \int_{0}^{\pi} |\sqrt{2} \sin x + 2\cos x| dx \).
Answer: \( I = \int_{0}^{\pi} |\sqrt{2} \sin x + 2\cos x| dx \)
\( \sqrt{2} \sin x + 2\cos x = 0 \)
\( \tan x = -\sqrt{2} \)
\( = \int_{0}^{\pi - \tan^{-1}\sqrt{2}} (\sqrt{2}\sin x + 2\cos x)dx - \int_{\pi - \tan^{-1}\sqrt{2}}^{\pi} (\sqrt{2}\sin x + 2\cos x)dx \)
\( = [-\sqrt{2} \cos x + 2 \sin x]_{0}^{\pi - \tan^{-1}\sqrt{2}} - [-\sqrt{2} \cos x + 2 \sin x]_{\pi - \tan^{-1}\sqrt{2}}^{\pi} \)
\( = \sqrt{2} \cos(\tan^{-1}\sqrt{2}) + \sin(\tan^{-1}\sqrt{2}) - (-\sqrt{2} + 0) + \sqrt{2} \cos x + 2 \sin x + \sqrt{2} \cos(\tan^{-1}\sqrt{2}) + 2 \sin(\tan^{-1}\sqrt{2}) \)
\( = \sqrt{2}\frac{1}{\sqrt{3}} + 2\frac{\sqrt{2}}{\sqrt{3}} + \sqrt{2} - \sqrt{2} + 0 + \frac{\sqrt{2}}{\sqrt{3}} + 2\frac{\sqrt{2}}{\sqrt{3}} \)
\( = 6\sqrt{\frac{2}{3}} = 2\sqrt{6} \)

Question. Evaluate the integral \( \int_{3}^{5} (\sqrt{x + 2\sqrt{2x - 4}} + \sqrt{x - 2\sqrt{2x - 4}}) dx \)
Answer: \( I = \int_{3}^{5} (\sqrt{x + 2\sqrt{2x - 4}} + \sqrt{x - 2\sqrt{2x - 4}}) dx \)
\( f^2(x) = x + 2\sqrt{2x - 4} + x - 2\sqrt{2x - 4} + 2\sqrt{x^2 - 4(2x - 4)} \)
\( f^2(x) = 2x + 2\sqrt{x^2 - 8x + 16} \)
\( f^2(x) = 2x + 2(x - 4) = 4x - 8 \)
\( f(x) = 2\sqrt{x - 2} \)
\( I = \int_{3}^{5} 2\sqrt{x - 2} dx = 2\left[\frac{(x - 2)^{3/2}}{3/2}\right]_{3}^{5} \)
\( = \frac{4}{3} [(3)^{3/2} - 1] = \frac{4}{3} [3\sqrt{3} - 1] = 4\sqrt{3} - \frac{4}{3} \)

Question. If \( P = \int_{0}^{\infty} \frac{x^2}{1 + x^4} dx \); \( Q = \int_{0}^{\infty} \frac{x dx}{1 + x^4} \) and \( R = \int_{0}^{\infty} \frac{dx}{1 + x^4} \)
then prove that
(a) \( Q = \frac{\pi}{4} \)
(b) \( P = R \)
(c) \( P - \sqrt{2} Q + R = \frac{\pi}{2\sqrt{2}} \)
Answer: \( P = \int_{0}^{\infty} \frac{x^2}{1 + x^4} dx \)
\( R = \int_{0}^{\infty} \frac{dx}{1 + x^4} \quad \dots(1) \)
pout \( x = \frac{1}{t} \Rightarrow dx = -\frac{dt}{t^2} \)
\( = -\int_{\infty}^{0} \frac{t^2}{1 + t^4} dt = \int_{0}^{\infty} \frac{t^2}{1 + t^4} dt = P \)
\( P = R \)
\( I = \int_{0}^{\infty} \frac{x^2 dx}{1 + x^4} \quad \dots(2) \)
add (1) + (2)
\( 2I = \int_{0}^{\infty} \frac{x^2}{x^4 + 1} dx + \int_{0}^{\infty} \frac{1}{x^4 + 1} dx = \int_{0}^{\infty} \left( \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \right) dx \)
Put \( x - \frac{1}{x} = t \Rightarrow \left(1 + \frac{1}{x}\right) dx = dt \)
\( = \int \frac{dt^2}{t^2(\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1} \frac{x - \frac{1}{x}}{\sqrt{2}} \Big|_{0}^{\infty} = \frac{1}{\sqrt{2}} \left[ \frac{\pi}{2} \right] \)
\( a = \int_{0}^{\infty} \frac{x dx}{1 + x^4} \)
Put \( x^2 = t \Rightarrow x dx = \frac{dt}{2} \)
\( = \frac{1}{2} \int_{0}^{\infty} \frac{dt}{1 + t^2} = \frac{1}{2} [\tan^{-1} t]_{0}^{\infty} = \frac{\pi}{4} \)
\( P - \sqrt{2} \cdot 2 + R = \frac{\pi}{2\sqrt{2}} \)

Question. \( \int_{0}^{1} \frac{x^4(1 - x)^4}{1 + x^2} dx \)
Answer: \( I = \int_{0}^{1} \left( \frac{x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{x^2 + 1}}{1} \right) dx \)
\( I = \left[ \frac{x^7}{7} - \frac{4x^6}{6} + x^5 - \frac{4x^3}{3} + 4x - 4\tan^{-1} x \right]_{0}^{1} \)
\( I = \frac{22}{7} - \pi \)

Question. \( \int_{0}^{1} \frac{x^2 \cdot \ln x}{\sqrt{1 - x^2}} dx \)
Answer: \( I = \int_{0}^{1} \frac{x^2 \cdot \ln x}{\sqrt{1 - x^2}} dx \)
put \( x = \sin\theta, dx = \cos\theta d\theta \)
\( I = \int_{0}^{\pi/2} \sin^2\theta \ln \sin\theta d\theta \)
\( I = \int_{0}^{\pi/2} \left( \frac{1 - \cos 2\theta}{2} \right) \ln \sin\theta d\theta \)
\( = \frac{1}{2} \int_{0}^{\pi/2} \ln \sin\theta d\theta - \frac{1}{2} \int_{0}^{\pi/2} \cos 2\theta \ln \sin\theta d\theta \)
\( = \frac{\pi}{8} (1 - \ln 4) \)

Question. \( \int_{-2\sqrt{2}}^{2} \frac{x^2 - x}{\sqrt{x^2 + 4}} dx \)
Answer: \( I = \int_{-2\sqrt{2}}^{2} \frac{x^2 dx}{\sqrt{x^2 + 4}} - \int_{-2\sqrt{2}}^{2} \frac{x dx}{\sqrt{x^2 + 4}} \to 0 \text{ as it is an odd function} \)
\( = 2 \int_{3}^{2} \frac{x^2 dx}{\sqrt{x^2 + 4}} = 2 \int_{0}^{2} \frac{x^2 + 4 - 4}{\sqrt{x^2 + 4}} dx \)
\( = 2 \int_{0}^{2} \sqrt{x^2 + 4} dx - 8 \int_{0}^{2} \frac{dx}{\sqrt{x^2 + 4}} \)
\( = 2\left[\frac{x}{2} \sqrt{x^2 + 4} + 2\ln(x + \sqrt{x^2 + 4})\right]_{0}^{2} - 8 \ln(x + \sqrt{x^2 + 4})\big|_{0}^{2} \)
\( = 4\sqrt{2} - 4\ln(\sqrt{2} + 1) \)

Question. \( \int_{0}^{\sqrt{3}} \sin^{-1} \frac{2x}{1 + x^2} dx \)
Answer: \( I = \int_{0}^{1} 2\tan^{-1} x dx + \int_{1}^{\sqrt{3}} (\pi - 2\tan^{-1} x) dx \)
using by parts
\( = \frac{\pi\sqrt{3}}{3} \)

Question. \( \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin\left(\frac{\pi}{4} + x\right)} dx \)
Answer: \( I = \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin\left(\frac{\pi}{4} + x\right)} dx \)
\( I = \sqrt{2} \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin x + \cos x} dx \)
by king's property
\( I = \sqrt{2} \int_{0}^{\pi/2} \frac{a \cos x + b \sin x}{\sin x + \cos \pi} dx \)
\( 2I = \sqrt{2} \int_{0}^{\pi/2} \frac{a \cos x + b \sin x}{\sin x + \cos x} dx \)
\( I = \frac{(a + b)\pi}{4\sqrt{2}} \)

Question. \( \int_{0}^{2\pi} \frac{dx}{2 + \sin 2x} \)
Answer: \( I = \frac{1}{2} \int_{0}^{4\pi} \frac{dx}{2 + \sin x} \quad \dots(1) \)
\( I = \frac{1}{2} \int_{0}^{4\pi} \frac{dx}{2 - \sin x} \quad \dots(2) \)
\( 2I = \frac{1}{2} \int_{0}^{4\pi} \frac{4}{4 - \sin^2 x} dx \)
\( I = 4 \int_{0}^{\pi} \frac{dx}{4 - \sin^2 x} \Rightarrow I = 8 \int_{0}^{\pi/2} \frac{dx}{4 - \sin^2 x} \)
\( I = 8 \int_{0}^{\pi/2} \frac{\sec^2 x dx}{4 + 3\tan^2 x} \Rightarrow I = 8 \int_{0}^{\infty} \frac{dt}{3t^2 + 4} = \frac{2\pi}{\sqrt{3}} \)

Advanced Subjective Questions

Question. Evaluate : \( \int_{0}^{1} e^{n \tan^{-1} x} \cdot \sin^{-1}(\cos x) dx \).
Answer: \( \int_{0}^{1} e^{n \tan^{-1} x} \sin^{-1}(\cos x) dx \) \( = \int_{0}^{1} \tan^{-1} x \sin^{-1}(\sin(\frac{\pi}{2} - x)) dx \) \( = \int_{0}^{1} (\frac{\pi}{2} - x) e^{n \tan^{-1} x} dx \) \( = \frac{\pi}{2} \int_{0}^{1} e^{n \tan^{-1} x} dx - \int_{0}^{1} x e^{n \tan^{-1} x} dx \) use by parts

Question. If the derivative of \( f(x) \) w.r.to \( x \) is \( \frac{\cos x}{f(x)} \) then show that \( f(x) \) is a periodic function.
Answer: \( \frac{d}{dx} (f(x)) = \frac{\cos x}{f(x)} \) \( \int f(x) \cdot df(x) = \int \cos x dx \) \( \frac{f^{2}(x)}{2} = \sin x + C \) \( f(x) = \sqrt{2 \sin x + C} \) so \( f(x) \) is periodic

Question. Find the range of the function, \( f(x) = \int_{-1}^{1} \frac{\sin x dt}{1 - 2t \cos x + t^{2}} \).
Answer: \( f(x) = \int_{-1}^{1} \frac{\sin x dt}{\sin^{2} x + (t - \cos x)^{2}} \) \( = \frac{\sin x}{\sin x} \left[ \tan^{-1} \left( \frac{t - \cos x}{\sin x} \right) \right]_{-1}^{1} \) \( = \tan^{-1} (\tan x/2) + \tan^{-1} (\cot x/2) \) Case-1 : \( 0 < x < \pi \) \( f(x) = \pi/2 \) Case-2 : \( \pi < x < 2\pi \) \( f(x) = -\pi/2 \) so range = \( \left\{ -\frac{\pi}{2}, \frac{\pi}{2} \right\} \)

Question. A function f is defined in [-1, 1] as \( f'(x) = 2 x \sin \frac{1}{x} - \cos \frac{1}{x} ; x \ne 0 ; f(0) = 0 ; f(1/\pi) = 0 \). Discuss the continuity and derivability of f at \( x = 0 \).
Answer: \( f'(x) = 2x \sin 1/x - \cos 1/x \) integrating by parts \( f(x) = \sin 1/x \cdot x^{2} - \int \cos \left( \frac{1}{x} \right) \left( -\frac{1}{x^{2}} \right) x^{2} dx - \int \cos \frac{1}{x} dx + c \) \( f(x) = x^{2} \sin \frac{1}{x} + c \) ; \( c = 0 \) \( f(x) = x^{2} \sin \frac{1}{x} \) continuous & differentiable at x = 0.

Question. Let \( f(x) = \begin{cases} -1 & \text{if } -2 \le x \le 0 \\ |x - 1| & \text{if } 0 < x \le 2 \end{cases} \) and \( g(x) = \int_{-2}^{x} f(t) dt \). Test the continuity and differentiability of \( g(x) \) in (-2, 2).
Answer: \( g(x) = \int_{-2}^{x} f(t) dt \) Taking different integrals \( g(x) = \begin{cases} -(x + 2) & ; -2 \le x \le 0 \\ -2 + x - \frac{x^{2}}{2} & ; 0 < x < 1 \\ \frac{x^{2}}{2} - x - 1 & ; 1 \le x \le 2 \end{cases} \) Not differentiable at x = 0.

Question. Prove the inequalities
(a) \( \frac{\pi}{6} < \int_{0}^{1} \frac{dx}{\sqrt{4 - x^{2} - x^{3}}} < \frac{\pi \sqrt{2}}{8} \)
(b) \( 2 e^{-1/4} < \int_{0}^{2} e^{x^{2} - x} dx < 2e^{2} \)
(c) \( a < \int_{0}^{2\pi} \frac{dx}{10 + 3\cos x} < b \) then find a & b.
(d) \( \frac{1}{2} \le \int_{0}^{2} \frac{dx}{2 + x^{2}} \le \frac{5}{6} \)
Answer: (a) \( 0 < x^{3} < x^{2} \) \( -2x^{2} < -(x^{2} + x^{3}) < -x^{2} \) \( 4 - 2x^{2} < 4 - x^{2} - x^{3} < 4 - x^{2} \) \( \implies \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2}}} dx < \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2} - x^{3}}} dx < \int_{0}^{1} \frac{1}{\sqrt{4 - 2x^{2}}} dx \) \( \implies \left. \sin^{-1} \left( \frac{x}{2} \right) \right|_{0}^{1} < \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2} - x^{3}}} dx < \left. \frac{1}{\sqrt{2}} \sin^{-1} \frac{x}{\sqrt{2}} \right|_{0}^{1} \) \( \implies \frac{\pi}{6} < \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2} - x^{3}}} dx < \frac{\pi}{4\sqrt{2}} \) (b) \( f(x) = e^{x^{2} - x} \) \( f'(x) = e^{x^{2} - x} (2x - 1) = 0 \implies x = \frac{1}{2} \) \( e^{-1/4} (2 - 0) < I < e^{2} (2 - 0) \) \( 2 e^{-1/4} < I < 2e^{2} \) (c) \( -1 < \cos x < 1 \) \( -3 < 3 \cos x < 3 \) \( 7 < 10 + 3\cos x < 13 \) \( \frac{1}{13} < \frac{1}{10 + 3\cos x} < \frac{1}{7} \) \( \frac{1}{13} \int_{0}^{2\pi} dx < I < \frac{1}{7} \int_{0}^{2\pi} dx \) \( \frac{2\pi}{13} < I < \frac{2\pi}{7} \) (d) \( I = \int_{0}^{2} \frac{dx}{2 + x^{2}} = \left. \frac{1}{\sqrt{2}} \tan^{-1} \frac{x}{\sqrt{2}} \right|_{0}^{2} \) \( = \frac{1}{\sqrt{2}} \tan^{-1} \sqrt{2} \) \( \cong \frac{\pi}{4\sqrt{2}} \cong 0.555 \) always lie in the given interval.

Question. If \( y = \frac{1}{a} \int_{0}^{x} f(t) \sin a(x - t) dt \) then prove that \( \frac{d^{2}y}{dx^{2}} + a^{2}y = f(x) \).
Answer: \( y = \frac{1}{a} \int_{0}^{x} f(t) [\sin ax \cos at - \sin at \cos ax] dt \) \( = \frac{\sin ax}{a} \int_{0}^{x} f(t) \cos at dt - \frac{\cos ax}{a} \int_{0}^{x} f(t) \sin at dt \) \( \implies \frac{dy}{dx} = \frac{1}{a} \left\{ a \cos ax \int_{0}^{x} f(t) \cos at dt + \sin ax (f(x) \cos ax) - a \sin ax \int_{0}^{x} f(t) \sin at dt - \cos ax (f(x) \sin ax) \right\} \) \( = \cos ax \int_{0}^{x} f(t) \cos at dt + \sin ax \int_{0}^{x} f(t) \sin at dt \) \( \implies \frac{d^{2}y}{dx^{2}} = -a \sin ax \int_{0}^{x} f(t) \cos at dt + a \cos ax (f(x) \cos ax) + a \cos ax \int_{0}^{x} f(t) \sin at dt + a \sin ax (f(x) \sin ax) \) \( [f(x) \sin ax] = - a^{2}y + f(x) \) \( \implies \frac{d^{2}y}{dx^{2}} + a^{2}y = f(x) \) HP.

Question. If \( y = x^{\int_{1}^{x} \ln t dt} \), find \( \frac{dy}{dx} \) at \( x = e \).
Answer: \( y = x^{\int_{1}^{x} \ln t dt} = x^{x \ln x - x + 1} \) \( \frac{dy}{dx} = y \left( \frac{1}{x} (x \ln x - x + 1) + \ln x (\ln x) \right) \) at x = e ; \( \frac{dy}{dx} = 1 + e \)

Question. If \( f(x) = x + \int_{0}^{1} [xy^{2} + x^{2}y] f(y) dy \) where x and y are independent variable, Find \( f(x) \).
Answer: \( f(x) = x + x \int_{0}^{1} y^{2} f(y) dy + x^{2} \int_{0}^{1} y f(y) dy \) Let \( A = \int_{0}^{1} y^{2} f(y) dy \) & \( B = \int_{0}^{1} y f(y) dy \) Now \( f(x) = x + Ax + Bx^{2} \) \( f(y) = y + Ay + By^{2} \) \( A = \int_{0}^{1} y^{2} (y + Ay + By^{2}) dy \) \( A = \int_{0}^{1} (y^{3} + Ay^{3} + By^{4}) dy \) \( A = \frac{1}{4} + \frac{A}{4} + \frac{B}{5} \implies \frac{3A}{4} - \frac{B}{5} = \frac{1}{4} \dots(1) \) \( B = \int_{0}^{1} y (y + Ay + By^{2}) dy \) \( \frac{A}{3} - \frac{3B}{4} = - \frac{1}{3} \dots(2) \) Solve (1) & (2) \( A = \frac{61}{119} \) & \( B = \frac{80}{119} \) Put the value of A & B in f(x)

Question. (a) Let \( g(x) = x^{c} \cdot e^{2x} \) & let \( f(x) = \int_{0}^{x} e^{2t} \cdot (3t^{2} + 1)^{1/2} dt \). For a certain value of 'c', the limit of \( \frac{f'(x)}{g'(x)} \) as \( x \to \infty \) is finite and non zero. Determine the value of 'c' and the limit.
(b) Find the constants 'a' (a > 0) and 'b' such that, \( \lim_{x \to 0} \frac{\int_{0}^{x} \frac{t^{2}}{\sqrt{a + t}} dt}{bx - \sin x} = 1 \).
Answer: (a) \( L = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{e^{2x}(3x^{2} + 1)^{1/2}}{e^{2x} x^{c} \left[ \frac{c}{x} + 2 \right]} \) for the existance of limit \( c - 1 = 0 \implies c = 1 \) & \( L = \frac{\sqrt{3}}{2} \) (b) \( \lim_{x \to 0} \frac{\int_{0}^{x} \frac{t^{2} dt}{\sqrt{a + t}}}{bx - \sin x} \) (\( \frac{0}{0} \) form) using L' Hospital \( L = \lim_{x \to 0} \frac{x^{2}}{(b - \cos x)\sqrt{a + x}} \) for the existance of limit : \( b = 1 \) & \( L = \lim_{x \to 0} \frac{x^{2}}{(1 - \cos x)} \frac{1}{\sqrt{a + x}} = 1 \implies a = 4 \)

Question. Evaluate : \( \lim_{x \to +\infty} \frac{d}{dx} \int_{2 \sin \frac{1}{x}}^{3\sqrt{x}} \frac{3t^{4} + 1}{(t - 3)(t^{2} + 3)} dt \)
Answer: Using Libnitz theorem \( L = \frac{3(3\sqrt{x})^{4} + 1}{(3\sqrt{x} - 3)(9x + 3)} \cdot \frac{3}{2\sqrt{x}} \) divide & multiply by \( x^{2} \) & \( L = 13.5 \)

Question. Supose \( g(x) \) is the inverse of \( f(x) \) and \( f(x) \) has a domain \( x \in [a, b] \). Given \( f(a) = \alpha \) and \( f(b) = \beta \), then find the value of \( \int_{a}^{b} f(x) dx + \int_{\alpha}^{\beta} g(y) dy \) in terms of a, b, \( \alpha \) and \( \beta \).
Answer: \( \alpha = f(a) \implies a = g(\alpha) \) \( \beta = f(b) \implies b = g(\beta) \) Let \( y = f(x) \) & \( x = g(y) \because g \) is the inverse of f \( dy = f'(x) dx \) \( \int_{a}^{b} f(x) dx + \int_{\alpha}^{\beta} g(y) dy = \int_{a}^{b} f(x) dx + \int_{a}^{b} x \cdot f'(x) dx \) \( = \int_{a}^{b} \{f(x) + x f'(x)\} dx = [x f(x)]_{a}^{b} \) \( = b f(b) - a f(a) = b\beta - a\alpha \)

Question. Evaluate
(a) \( \lim_{n \to \infty} \left[ \left( 1 + \frac{1^{2}}{n^{2}} \right) \left( 1 + \frac{2^{2}}{n^{2}} \right) \left( 1 + \frac{3^{2}}{n^{2}} \right) \dots \left( 1 + \frac{n^{2}}{n^{2}} \right) \right]^{1/n} \)
(b) \( \lim_{n \to \infty} \frac{1}{n} \left[ \frac{1}{n + 1} + \frac{2}{n + 2} + \dots + \frac{3n}{4n} \right] \)
(c) \( \lim_{n \to \infty} \left[ \frac{n!}{n^{n}} \right]^{1/n} \)
(d) For positive integers n, let
\( A_{n} = \frac{1}{n} \{(n + 1) + (n + 2) + \dots (n + n)\} \),
\( B_{n} = \{(n + 1) (n + 2) \dots (n + n)\}^{1/n} \).
If \( \lim_{n \to \infty} \frac{A_{n}}{B_{n}} = \frac{ae}{b} \) where a, b \( \in \mathbb{N} \) and relatively prime find the value of (a + b).
Answer: (a) Taking log : \( \ln y = \lim_{n \to \infty} \frac{1}{n} \left\{ \ln \left( 1 + \frac{1^{2}}{n^{2}} \right) + \ln \left( 1 + \frac{2^{2}}{n^{2}} \right) + \dots + \ln \left( 1 + \frac{n^{2}}{n^{2}} \right) \right\} \) \( \ln y = \int_{0}^{1} \ln(1 + x^{2}) dx \) \( y = 2 e^{1/2 (\pi - 4)} \) (b) \( \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{3n} \frac{r}{n + r} \) \( \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{3n} \frac{r / n}{1 + (r / n)} \) \( I = \int_{0}^{3} \frac{x}{1 + x} dx = \int_{0}^{3} \frac{x + 1 - 1}{x + 1} dx \) \( = \int_{0}^{3} dx - \int_{0}^{3} \frac{dx}{x + 1} = [x]_{0}^{3} - [\ln(1 + x)]_{0}^{3} \) \( I = 3 - \ln 4 \) (c) Let \( y = \lim_{n \to \infty} \left( \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \dots \frac{n}{n} \right)^{1/n} \) \( \ln y = \lim_{n \to \infty} \frac{1}{n} \left[ \ln \left( \frac{1}{n} \right) + \ln \left( \frac{2}{n} \right) + \dots + \ln \left( \frac{n}{n} \right) \right] \) \( = \int_{0}^{1} \ln x dx = -1 \implies y = e^{-1} \) (d) \( \lim_{n \to \infty} \frac{A_{n}}{B_{n}} = \frac{n + \frac{n(n + 1)}{2n}}{\left( (n + 1)(n + 2) \dots (n + n) \right)^{1/n}} \) \( = \frac{n + \frac{n + 1}{2}}{n \left( \left( 1 + \frac{1}{n} \right) \left( 1 + \frac{2}{n} \right) \dots \left( 1 + \frac{n}{n} \right) \right)^{1/n}} \) \( \ln y = \lim_{n \to \infty} \left\{ \ln \left( \frac{3}{2} + \frac{1}{2n} \right) - \frac{1}{n} \left( \ln \left( 1 + \frac{1}{n} \right) + \ln \left( 1 + \frac{2}{n} \right) \dots \ln \left( 1 + \frac{n}{n} \right) \right) \right\} \) \( = \ln \left( \frac{3}{2} \right) - \int_{0}^{1} \ln(1 + x) dx \)

Question. Prove that \( \sin x + \sin 3x + \sin 5x + \dots + \sin (2k - 1) x = \frac{\sin^{2} kx}{\sin x} \), \( k \in \mathbb{N} \) and hence prove that,
\( \int_{0}^{\pi/2} \frac{\sin^{2} kx}{\sin x} dx = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots + \frac{1}{2k - 1} \).
Answer: \( \sin x + \sin 3x + \dots + \sin (2k - 1) x \) \( = \frac{\sin \left( \frac{2x}{2} \cdot k \right)}{\sin \left( \frac{2x}{2} \right)} \times \sin \left( \frac{x + (2k - 1) x}{2} \right) = \frac{\sin^{2} kx}{\sin x}, k \in \mathbb{N} \) \( \int_{0}^{\pi/2} \frac{\sin^{2} kx}{\sin x} dx = \int_{0}^{\pi/2} \sin x dx + \int_{0}^{\pi/2} \sin 3x dx \dots + \int_{0}^{\pi/2} \sin (2k - 1) dx \) \( = 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2k - 1} \)

Question. Solve the equation for y as a function of x, satisfying \( x \cdot \int_{0}^{x} y(t) dt = (x + 1) \int_{0}^{x} t \cdot y(t) dt \), where \( x > 0 \), given \( y(1) = 1 \).
Answer: \( x \cdot \int_{0}^{x} y(t) dt = (x + 1) \int_{0}^{x} t \cdot y(t) dt \) Differentiating w.r.t. 'x' \( \int_{0}^{x} y(t) dt + x y(x) = \int_{0}^{x} t y(t) dt + (x + 1)x y(x) \) \( \implies \int_{0}^{x} (1 - t) y(t) dt = x^{2} y(x) \) again differentiating w.r.t. 'x' \( y'(x) = y(x) \left( \frac{1 - 3x}{x^{2}} \right) \) Integrate & put y(1) = 1 \( y(x) = \frac{e}{x^{3}} e^{-1/x} \)

Question. Prove that
(a) \( I_{m, n} = \int_{0}^{1} x^{m} \cdot (1 - x)^{n} dx = \frac{m! n!}{(m + n + 1)!}, m, n \in \mathbb{N}. \)
(b) \( I_{m, n} = \int_{0}^{1} x^{m} \cdot (\ln x)^{n} dx = (-1)^{n} \frac{n!}{(m + 1)^{n + 1}}, m, n \in \mathbb{N}. \)
Answer: (a) \( I_{m, n} = \int_{0}^{1} x^{m} (1 - x)^{n} dx \) \( I_{m, n} = \left[ \frac{x^{m+1}}{m+1} (1 - x)^{n} \right]_{0}^{1} + \frac{n}{m+1} \int_{0}^{1} x^{m+1} (1 - x)^{n-1} dx \) & so on .... so \( I_{m, n} = \frac{m! n!}{(m + n + 1)!}, m, n \in \mathbb{N} \) (b) \( I_{m, n} = \int_{0}^{1} x^{m} (\ln x)^{n} dx \)

Question. Find a positive real valued continuously differentiable functions f on the real line such that for all x
\( f^{2}(x) = \int_{0}^{x} ((f(t))^{2} + (f'(t))^{2}) dt + e^{2} \)
Answer: \( f^{2}(x) = \int_{0}^{x} \{(f(t))^{2} + (f'(t))^{2}\} dt + e^{2} \) differentiate both the side w.r.t. x \( 2f(x) f'(x) = \{f(x)\}^{2} + \{f'(x)\}^{2} \cdot 1 + 0 \) \( \{f(x)\}^{2} - 2f(x) f'(x) + \{f'(x)\}^{2} = 0 \) \( \{f(x) - f'(x)\}^{2} = 0 \implies f'(x) = f(x) \) \( \implies \int \frac{f'(x) dx}{f(x)} = \int dx \) \( \ln f(x) = x + c \implies f(x) = e^{x + c} \dots(1) \) \( f(x) = e^{x} \cdot e^{c} \) Now put x = 0 in given equation \( f^{2}(0) = c^{2} \implies e^{c} = e^{1} \implies c = 1 \) put c = 1 in equation (1) \( f(x) = e^{x + 1} \)

Question. Let \( f(x) \) be a continuously differentiable function then prove that, \( \int_{1}^{x} [t] f'(t) dt = [x] \cdot f(x) - \sum_{k=1}^{[x]} f(k) \)
(where \( [ * ] \) denotes the greatest integer function and \( x > 1 \))
Answer: \( \int_{1}^{x} [t] f'(t) dt = \int_{1}^{2} [t] f'(t) dt + \int_{2}^{3} [t] f'(t) dt + \dots + \int_{n}^{n+f} [t] f'(t) dt \) where \( n \in \mathbb{N} \) & \( 0 \le f < 1 \) \( = \int_{1}^{2} f'(t) dt + 2 \int_{2}^{3} f'(t) dt + 3 \int_{3}^{4} f'(t) dt + \dots + n \int_{n}^{n+f} f'(t) dt \) \( = [f(t)]_{1}^{2} + 2 [f(t)]_{2}^{3} + 3 [f(t)]_{3}^{4} + \dots + n [f(t)]_{n}^{n+f} \) \( = - f(1) - f(2) - f(3) \dots f(n) + n f(n + f) \) \( = - [f(1) + f(2) + \dots + f(n)] + [x] f(x) \) \( = [x] f(x) - \sum_{k=1}^{[x]} f(k) \)

Question. Let \( f(x) = \int_{-1}^{x} \sqrt{4 + t^{2}} dt \) and \( G(x) = \int_{x}^{1} \sqrt{4 + t^{2}} dt \) then compute the value of \( (FG)' (0) \) where dash denotes the derivative.
Answer: \( (FG)' = F'G + FG' \) \( = (\sqrt{4 + x^{2}}) \int_{x}^{1} \sqrt{4 + t^{2}} dt + (-\sqrt{4 + x^{2}}) \int_{-1}^{x} \sqrt{4 + t^{2}} dt \) at x = 0 \( (FG)' = 0 \)

Question. Show that for a continuously thrice differentiable function \( f(x) \)
\( f(x) - f(0) = x f'(0) + \frac{f''(0) \cdot x^{2}}{2} + \frac{1}{2} \int_{0}^{x} f'''(t)(x - t)^{2} dt \)
Answer: \( \because \int_{0}^{x} f'''(t) (x - t)^{2} dt \) \( = [(x - t)^{2} f''(t)]_{0}^{x} + 2 \int_{0}^{x} (x - t) f''(t) dt \) \( = [-x^{2} f''(0)] + 2 \left[ \left[ (x - t) f'(t) \right]_{0}^{x} + \int_{0}^{x} f'(t) dt \right] \) \( = -x^{2} f''(0) - 2x f'(0) + 2f(x) - 2f(0) \) Replace this value in R.H.S. \( = x f'(0) + \frac{x^{2} f''(0)}{2} - \frac{x^{2}}{2} f''(0) - x f'(0) + f(x) - f(0) \) \( = f(x) - f(0) = \text{L.H.S.} \) Hence proved

Question. Let f and g be function that are differentiable for all real numbers x and that have the following properties
(i) \( f'(x) = f(x) - g(x) \)
(ii) \( g'(x) = g(x) - f(x) \)
(iii) \( f(0) = 5 \)
(iv) \( g(0) = 1 \)
(a) Prove that \( f(x) + g(x) = 6 \) for all x.
(b) Find \( f(x) \) and \( g(x) \).
Answer: (a) \( f'(0) = 4, g'(0) = -4 \) \( f'(x) = -g'(x) \) Integrating \( \implies f(x) = -g(x) + c \) \( f(x) + g(x) = c \) at x = 0, c = 6 so \( f(x) + g(x) = 6 \) (b) Now \( f'(x) = f(x) - g(x) = 2f(x) - 6 \) Integrating & using f(0) = 5 \( f(x) = 3 + 2e^{2x} \) & \( g(x) = 3 - 2e^{2x} \)