Question

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

Answer 1

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

add & subtract $2{\sin }^{2}x{\cos }^{2}x$

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

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

$={\int }_0^{\pi /4}\frac{\sin x\cos xdx}{1-\frac{{\sin }^{2}2x}{2}}={\int }_0^{\pi /4}\frac{2\sin x\cos xdx}{2-{\sin }^{2}2x}$

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

Let $t=\cos 2x$

$dt=-2\sin 2xdx$

$\left|\begin{array}{cc}x& t=\cos 2x\\ 0& 1\\ \pi /4& 0\end{array}\right|$

$=\frac{-1}{2}{\int }_1^0\frac{dt}{1+t^2}$

$\Rightarrow \frac{-1}{2}\left[{\tan }^{-1}\right(t){]}_1^0$

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

$=\frac{-1}{2}[0-\frac{\pi }{4}]=\frac{\pi }{8}$.