ISI-2014-07

Question

The value of the integral ${\LARGE \int} _{0}^{\pi}\dfrac{x}{1+sin^2x}dx$ is

• A. $2\sqrt2\pi^2$
• B. $\dfrac{\pi^2}{2\sqrt2}$
• C. $\dfrac{\pi^2}{\sqrt2}$
• D. $\sqrt2\pi^2$

Comments

Is the answer $\dfrac{\pi^2}{2\sqrt2}$ ?

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$\dfrac{\pi^2}{2\sqrt{2}}$

Answer 1

Assuming , $I = {\LARGE \int} _{0}^{\pi}\dfrac{x}{1+\sin^2x}dx$

$\qquad ={\LARGE \int} _{0}^{\pi}\dfrac{(\pi-x)}{1+\sin^2(\pi-x)}dx$

$\qquad =\pi{\LARGE \int} _{0}^{\pi}\dfrac{dx}{1+\sin^2x}-{\LARGE \int} _{0}^{\pi}\dfrac{dx}{1+\sin^2x}dx$       $\Big[$$\color{blue}{\text{We know that}}$ $\color{red}{\sin(\pi-x)=\sin x}\Big]$

∴ $I =\pi{\LARGE \int} _{0}^{\pi}\dfrac{dx}{1+\sin^2x}-I$

Or, $2I =\pi{\LARGE \int} _{0}^{\pi}\dfrac{dx}{1+\sin^2x}$

$\qquad = 2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{dx}{1+\sin^2x}$ $\qquad\qquad\Bigg[\color{red}{∵ {\LARGE \int} _{0}^{2a}f(x)dx=2{\LARGE \int} _{0}^{a}f(x)dx} \hspace{0.2cm}\color{blue}{\text{when}}\hspace{0.1cm}\color{red}{ f(a-x)dx  = f(x)}\Bigg]$

$\qquad = 2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{dx}{1+\dfrac{1}{\mathrm{cosec}^2x}}= 2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{\mathrm{cosec}^2x}{1+\mathrm{cosec}^2x}dx$

$\qquad=2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{\mathrm{cosec}^2x}{1+1+\cot^2x}dx$  $\qquad\qquad\bigg[\color{red}{∵\mathrm{cosec}^2x-\cot^2x=1}\bigg]$

$\qquad=2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{\mathrm{cosec}^2x}{2+\cot^2x}dx$

$\color{blue}{\text{Now, assuming,}}$ $\color{red}{\cot x=z}$

$\qquad\color{red}{∴\dfrac{d}{dz}(\cot x)= dz}$

$\qquad Or, \color{red}{-\mathrm{cosec}^2x.dx=dz}$

$\color{blue}{\&}$ 

$\color{red}{x}$	$\color{red}{0}$	$\color{red}{\dfrac{\pi}{2}}$
$\color{red}{z}$	$\color{red}{\infty}$	$\color{red}{0}$

$∴2\pi{\LARGE \int} _{0}^{\dfrac{\pi}{2}}\dfrac{\mathrm{cosec}^2x}{2+\cot^2x}dx = -2\pi{\LARGE \int} _{\infty}^{0}\dfrac{dz}{z^2+2} =2\pi{\LARGE \int} _{0}^{\infty}\dfrac{dz}{z^2+2} $ $\qquad\Bigg[\color{red}{∵{\LARGE \int_a^bdx= -\LARGE \int_a^bdx}}\Bigg]$

$= 2\pi{\LARGE \int} _{0}^{\infty}\dfrac{dz}{z^2+(\sqrt{2})^2}$

$∴2I = 2\pi{\LARGE \int} _{0}^{\infty}\dfrac{dz}{z^2+(\sqrt{2})^2}$

$Or,I = \pi{\LARGE \int} _{0}^{\infty}\dfrac{dz}{z^2+(\sqrt{2})^2}$

$\color{blue}{\text{Now, we know}}$ $\color{red}{\Large \int\dfrac{dx}{x^2+a^2}= \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} +c} \hspace{0.2cm} \color{blue}{where,\hspace{0.1cm}c=constant}$

$\color{green}{∴\pi{\LARGE \int} _{0}^{\infty}\dfrac{dz}{z^2+(\sqrt{2})^2}=\pi\bigg[\dfrac{1}{\sqrt2}\tan^{-1}\dfrac{x}{\sqrt2}\bigg]_0^\infty= \pi.\dfrac{1}{\sqrt2}.\dfrac{\pi}{2}=\dfrac{\pi^2}{2\sqrt2}}$

Comments

wonderful explanation !!!!