Question

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

Answer 1

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

$=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x\cos x}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}dx$

$=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x\cos x}{1-\frac{\pi }{2}2{\sin }^{2}x{\cos }^{2}x}dx$

$=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin x\cos x}{1-\frac{{\sin }^{2}2x}{2}}$

$=\underset{0}{\overset{\pi /4}{\int }}\frac{2\sin x\cos x}{2-{\sin }^{2}x2x}dx$

$=\underset{0}{\overset{\pi /4}{\int }}\frac{\sin 2x}{1+{\cos }^{2}x}dx$

Let $t=\cos 2x$

$dt=-2\sin 2xdt$

$=-\frac{1}{2}\underset{1}{\overset{0}{\int }}\frac{dt}{1+t^2}=-\frac{1}{2}{\left[{\tan }^{-1}t\right]}_1^0$

$=-\frac{1}{2}\left[{\tan }^{-1}0-{\tan }^{-1}1\right]=-\frac{1}{2}\left[0-\frac{\pi }{4}\right]$

$=\frac{\pi }{8}$