Question

$I=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)dx}{1+{\cos }^{2}x}$, Find the value of $[\frac{I}{{\pi }^2}]$ (where [.] is the greatest integer function)

Answer 1

$I=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)dx}{1+{\cos }^{2}x}=\underset{-\pi }{\overset{\pi }{\int }}\frac{2xdx}{1+{\cos }^{2}x}+\underset{-\pi }{\overset{\pi }{\int }}\frac{2x\left(\sin x\right)dx}{1+{\cos }^{2}x}$
$\because \underset{-\pi }{\overset{\pi }{\int }}\frac{2xdx}{1+{\cos }^{2}x}=0$ ($\because$ odd function) $\underset{-\pi }{\overset{\pi }{\int }}\frac{2x\left(\sin x\right)dx}{1+{\cos }^{2}x}=4\underset{0}{\overset{\pi }{\int }}\frac{x\left(\sin x\right)dx}{1+{\cos }^{2}x}$ ($\because$ Even function).
$\therefore I=4\underset{0}{\overset{\pi }{\int }}\frac{x\left(\sin x\right)dx}{1+{\cos }^{2}x}$
$I=4\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)\left(\sin x\right)dx}{1+{\cos }^{2}x}$
$I=4\underset{0}{\overset{\pi }{\int }}\frac{\pi \left(\sin x\right)dx}{1+{\cos }^{2}x}-I$
$I=4\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin xdx}{1+{\cos }^{2}x}-I$
$2I=4\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin xdx}{1+{\cos }^{2}x}$
Let $\cos x=t\Rightarrow -\sin xdx=dt$
$I=-2\pi \underset{1}{\overset{-1}{\int }}\frac{1}{1+t^2}dt=2\pi \underset{-1}{\overset{1}{\int }}\frac{1}{1+t^2}dt=4\pi \underset{0}{\overset{1}{\int }}\frac{1}{1+t^2}dt=4\pi [{\tan }^{-1}t{]}_0^1=4\pi \times \frac{\pi }{4}={\pi }^2$
Hence, $[\frac{I}{{\pi }^2}]=1$