Question

Evaluate ${\int }_0^{\pi }\frac{x{\sin }^{3}x}{1+{\cos }^{2}x}dx$.

Answer 1

$I={\int }_0^{\pi }\frac{xsi{n}^{3}x}{1+co{s}^{2}x}dx$
$={\int }_0^{\pi }\frac{(\pi -x)si{n}^3(\pi -x)}{1+co{s}^2(\pi -x)}dx={\int }_0^{\pi }\frac{(\pi -x)si{n}^{3}x}{1+co{s}^{2}x}dx$
Hence
$2I={\int }_0^{\pi }\frac{\pi si{n}^{3}x}{1+co{s}^{2}x}dx$
Or
$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{si{n}^{3}x}{1+co{s}^{2}x}dx$
Let $cosx=t$
$sinxdx=-dt$
Hence
$I=\frac{-\pi }{2}{\int }_0^{-1}\frac{(1-t^2)dt}{1+t^2}$
$=\frac{\pi }{2}{\int }_0^{-1}\frac{(t^2-1)dt}{t^2+1}$
$=\frac{\pi }{2}{\int }_0^{-1}\frac{(t^2+1-2)dt}{t^2+1}$
$=\frac{\pi }{2}{\int }_0^{-1}1-\frac{2}{t^2+1}dt$
$=\frac{\pi }{2}[t-2ta{n}^{-1}t{]}_0^{-1}$
$=\frac{\pi }{2}[-1-2ta{n}^{-1}(-1\left)\right]$
$=\frac{\pi }{2}[-1-2(\frac{-\pi }{4}\left)\right]$
$=\frac{\pi }{2}(\frac{\pi }{2}-1)$
$=\frac{{\pi }^2}{4}-\frac{\pi }{2}$