Question

The value of the integral $\frac{48}{\pi^{4}} \int\limits_{0}^{\pi}\left(\frac{3 \pi x^{2}}{2}-x^{3}\right) \frac{\sin x}{1+\cos ^{2} x} d x$ is equal to _______.

Answer 1

$I=\frac{48}{\pi^{4}} \int\limits_{0}^{\pi} x^{2}\left(\frac{3 \pi}{2}-x\right) \frac{\sin x}{1+\cos ^{2} x} d x ....$(1)
Apply king property
$I =\frac{48}{\pi^{4}} \int\limits_{0}^{\pi}(\pi- x )^{2}\left(\frac{\pi}{2}+ x \right) \frac{\sin x }{1+\cos ^{2} x } dx \ldots(2)$
$(1)+(2)$
$I =\frac{12}{\pi^{3}} \int\limits_{0}^{\pi} \frac{\sin x }{1+\cos ^{2} x }\left[\pi^{2}+(\pi-2) \cdot x \cdot(\pi-2 x )\right] dx \ldots(3)$
Apply king again
$I =\frac{12}{\pi^{3}} \int\limits_{0}^{\pi} \frac{\sin x }{1+\cos ^{2} x }\left[\pi^{2}+(\pi-2)(\pi- x )(2 x -\pi)\right] dx \ldots$ (4)
(3) $+(4)$
$I =\frac{6}{\pi^{2}} \int\limits_{0}^{\pi} \frac{\sin x }{1+\cos ^{2} x }[2 \pi+(\pi-2)(\pi-2 x )] dx \ldots(5)$
Apply king
$I=\frac{6}{\pi^{2}} \int\limits_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x}[2 \pi+(\pi-2)(2 x-\pi)] d x \ldots(6)$
(5) $+(6)$
$I =\frac{12}{\pi} \int\limits_{0}^{\pi} \frac{\sin x }{11 \cos ^{2} x } dx$
Let $\cos x = t \Rightarrow \sin xdx =- dt$
$I =\frac{12}{\pi} \int\limits_{1}^{-1} \frac{- dt }{1+ t ^{2}}=6$