Question

Solve ${\int }_0^{\pi /2}\frac{x\sin xcosx}{co{s}^{4}x+{\sin }^{4}x}dx$

Answer 1

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

Using , ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

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

We know, $\cos (\frac{\pi }{2}-x)=\sin x$ and $\sin (\frac{\pi }{2}-x)=\cos x$

Putting in $I$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)\cos x\sin x}{{\cos }^{4}x+{\sin }^{4}x}dx$

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

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

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

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

Dividing Numerator and Denominator with ${\cos }^{4}x$

$I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\frac{\sin x}{{\cos }^{3}x}}{\frac{{\sin }^{4}x}{{\cos }^{4}x}+1}dx$

$I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\tan x{\sec }^{2}x}{ta{n}^{4}x+1}dx$

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

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

When $x=0,t=0$ and $x=\frac{\pi }{2},t=\infty$

$I=\frac{\pi }{4}{\int }_0^{\infty }\frac{\frac{dt}{2}}{t^2+1}$

$I=\frac{\pi }{8}{\int }_0^{\infty }\frac{dt}{t^2+1}$

$I=\frac{\pi }{8}[{\tan }^{-1}t{]}_0^{\infty }$

$I=\frac{\pi }{8}[ta{n}^{-1}\infty -ta{n}^{-1}0]$

$I=\frac{\pi }{8}[\frac{\pi }{2}-0]$

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