Question

If the value of ${{\int }_0^{\pi }\left(\frac{x}{1+\sin x}\right)}^{2}dx=1,$ then the value of the integral ${\int }_0^{\pi }\left[\frac{2x^2{\cos }^{2}x/2}{{(1+\sin x)}^2}\right]dx$ is equal to

• A. $1+\pi (2-\pi )$
• B. $1-\pi (\pi -2)$
• C. $1+\pi (2+\pi )$
• D. $1-\pi (2+\pi )$

Answer 1

Answer: (A)

Given that, ${\int }_0^{\pi }{\left(\frac{x}{1+\sin x}\right)}^{2}dx=1$ ..(i)
Let $I={\int }_0^{\pi }\left[\frac{2x^2{\cos }^2(x/2)}{{(1+\sin x)}^2}\right]dx$
$\therefore$ $I={\int }_0^{\pi }\frac{x^2(1+\cos x)}{{(1+\sin x)}^2}dx$
$={\int }_0^{\pi }{\left(\frac{x}{1+\sin x}\right)}^{2}dx+{\int }_0^{\pi }\frac{x^2\cos x}{(1+\sin x)}dx$
$\Rightarrow$ $I=1+I_2$ ..(ii)
Where $I=1+I_2$
$\Rightarrow$ $I_2=-{\int }_0^{\pi }\frac{{(\pi -x)}^2\cos x}{{(1+\sin x)}^2}dx$
$\Rightarrow$ $2I_2={\int }_0^{\pi }\frac{(-{\pi }^2+2\pi x)}{{(1+\sin x)}^2}.dx$
$\Rightarrow$ $2I_2=-{\pi }^2{\int }_0^{\pi }\frac{\cos x}{{(1+\sin x)}^2}.dx$ $+2\pi {\int }_0^{\pi }\frac{x(\cos x)}{{(1+\sin x)}^2}dx$
$={\pi }^2{\left[\frac{1}{1+\sin x}\right]}_0^{\pi }+2\pi {\int }_0^{\pi }\frac{x\cos x}{{(1+\sin x)}^2}dx$
$\Rightarrow$ $I_2=\frac{{\pi }^2}{2}[1-1]+I_3$ ?..(iii) Where $I_3=\pi {\int }_0^{\pi }\frac{x\cos x}{{(1+\sin x)}^2}dx$
$=\pi \left[{\left\{-\frac{x}{1+\sin x)}\right\}}_0^{\pi }+{\int }_0^{\pi }\frac{1}{1+\sin x}dx\right]$
$=\pi \left[-\pi +{\int }_0^{\pi }\frac{{\sec }^{2}x/2}{{(1+\tan x/2)}^2}dx\right]$
$=\pi \left[-\pi -2{\left(\frac{1}{1+\tan \frac{x}{2}}\right)}_0^{\pi }\right]$
$=\pi (-\pi +2)$ On putting this value in Eq. (iii), we get $I_2=\pi (-\pi /2)$ On putting this value in Eq. (ii), we get $I=1+\pi (-\pi +2)$