Question

The value of the integral ${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$ is

• A. $\frac{{\pi }^2}{2}$
• B. $\frac{{\pi }^2}{4}$
• C. ${\pi }^2$
• D. None of these

Answer 1

Answer: (B)

$\frac{{\pi }^2}{4}$

${\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}\to I_1$
$={\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}$
$I_1+I_2=2I_1={\int }_0^{\pi }\frac{\pi sinx}{1+co{s}^{2}x}$
$I_1=\frac{\pi }{2}{\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}$
$cosx=t$
$sinxdx=dt$
$I_1=\frac{\pi }{2}{\int }_1^{-1}\frac{dt}{1+t^2}=\frac{\pi }{2}\times (\frac{\pi }{4}+\frac{\pi }{4})$
$=\frac{{\pi }^2}{4}$