Question

Solve${\int }_0^{\pi /2}\frac{xsinxcosx}{co{s}^{4}x+si{n}^{4}x}dx$

Answer 1

$I={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx.......\left(i\right)$

$I={\int }_0^{\pi /2}\frac{(\pi /2-x)\sin (\pi /2-x)\cos (\pi /2-x)}{{\cos }^4(\pi /2-x)+{\sin }^4(\pi /2-x)}$

$I={\int }_0^{\pi /2}\frac{(\pi /2-x)\cos x\sin x}{{\sin }^4+{\cos }^{4}x}dx......\left(ii\right)$

Ass $\left(i\right)$ & $\left(ii\right)$

$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\cos }^{4}x({\tan }^{4}x+1)}dx$

$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\tan x{\sec }^{2}x}{(1+{\tan }^{4}x)}dx$

Let ${\tan }^{2}x=t$

$2\tan x{\sec }^{2}xdx=dt$

$\tan x{\sec }^{2}xdx=\frac{dt}{2}$

$\int \frac{\tan x{\sec }^2}{1+{\tan }^{4}x}dx=\int \frac{dt}{21+t^2}$

$=\frac{1}{2}{\tan }^{-1}t$

$=\frac{1}{2}{\tan }^{-1}\left({\tan }^{2}x\right)$

$2I=\frac{\pi }{2}.\frac{1}{2}{\tan }^{-1}\left({\tan }^{2}x\right){\frac{}{}|}_0^{\pi /2}$

$2I=\frac{\pi }{4}[\frac{\pi }{2}-0]$

$I=\frac{{\pi }^2}{16}$

$\therefore {\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx=\frac{{\pi }^2}{16}$