Question

The value of ${\int }_0^{\pi }\frac{x\sin nx}{\sin x}dx$ equals, if $n$ is odd integer

• A. ${\pi }^2$
• B. $\frac{{\pi }^2}{2}$
• C. $\frac{3{\pi }^2}{2}$
• D. $2{\pi }^2$

Answer 1

Answer: (B)

$\frac{{\pi }^2}{2}$

$I={\int }_0^{\pi }\frac{x\sin nx}{\sin x}dx$
$I={\int }_0^{\pi }\frac{(\pi -x)\sin n(\pi -x)}{\sin x}(\because \sin (\pi -x)=\sin x)$
$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin nx}{\sin x}dx$
$I_m=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin (2m-1)x}{\sin x}dx...\left(i\right)(\because n=2m\pm 1)$
$\therefore I_{m+1}=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin (2m+1)x}{\sin x}dx$...(ii)
$\left(1\right)-\left(2\right)$ gives
$I_{m+1}-I_m=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin (2m+1)x-\sin (2m-1)x}{\sin x}dx$
$=\pi {\int }_0^{\pi }\cos 2mxdx=0\because \sin 2m\pi =0$
$I_{m+1}=I_m$ i.e$I_1=I_2=I_m$

$I_1={\int }_0^{\pi }xdx=\frac{{\pi }^2}{2}$