Question

Evaluate $\underset{0}{\overset{\pi /2}{\int }}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$.

Answer 1

Let $I=\underset{0}{\overset{\pi /2}{\int }}\frac{x\sin x\cdot \cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-x\right)\sin \left(\frac{\pi }{2}-x\right)\cdot \cos \left(\frac{\pi }{2}-x\right)}{{\sin }^4\left(\frac{\pi }{2}-x\right)+{\cos }^4\left(\frac{\pi }{2}-x\right)}dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-x\right)\cdot \sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
$\Rightarrow I\equiv \frac{\pi }{2}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx-\underset{0}{\overset{\pi /2}{\int }}\frac{x\sin x\cdot \cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$=\frac{\pi }{2}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x\cdot \cos x}{{\sin }^{4}x+{\cos }^{4}x}dx-I$
$\Rightarrow 2I=\frac{\pi }{2}\underset{0}{\overset{\pi /2}{\int }}\frac{\tan x\cdot {\sec }^{2}x}{{\tan }^{4}x+1}dx$
$\Rightarrow 2I=\frac{\pi }{2}\cdot \frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\frac{1}{1+{\left({\tan }^{2}x\right)}^2}d\left({\tan }^{2}x\right)$
$\Rightarrow 2I=\frac{\pi }{4}\cdot {\left[{\tan }^{-1}t\right]}_0^{\infty }=\frac{\pi }{4}\left({\tan }^{-1}\infty -{\tan }^{-1}0\right)$
[where, $t={\tan }^{2}x$ ]
$\Rightarrow I=\frac{{\pi }^2}{16}$