Question

Evaluate:${\int }_0^{\pi }\frac{4xsinx}{1+co{s}^{2}x}dx$

Answer 1

Let $I={\int }_0^{\pi }\frac{4xsinx}{1+co{s}^{2}x}dx=4{\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}dx......\left(i\right)$

Also,$I=4{\int }_0^{\pi }\frac{(\pi -x)sin(\pi -x)}{1+co{s}^2(\pi -x)}dx=4{\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx....\left(ii\right)$

Adding (i) and (ii)

$2I=4{\int }_0^{\pi }\frac{\pi sinx}{1+co{s}^{2}x}dx$

Put $cosx=t\Rightarrow -sinxdx=dt\Rightarrow sinxdx=-dt$

When $x=0,t=1,whenx=\pi ,t=-1$

$2I=4\pi {\int }_1^{-1}\frac{-dt}{1+t^2}=-4\pi {\int }_1^{-1}\frac{dt}{1+t^2}=4\pi {\int }_{-1}^1\frac{dt}{1+t^2}$

$=4\pi {\left({\tan }^{-1}x\right)}_{-1}^1=4\pi [{\tan }^{-1}1-ta{n}^{-1}(-1\left)\right]$

$2I=4\pi [\frac{\pi }{4}-\left(\frac{-\pi }{4}\right)]=4\pi \left(\frac{\pi }{2}\right)=2{\pi }^2$

$I={\pi }^2$