Question

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

Answer 1

Answer: 6

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