Question

${\int }_0^{\pi /4}\frac{\sin x\cos xdx}{{\cos }^{2}x+{\sin }^{4}x}$.

Answer 1

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

Divide numerator & denominator by ${\cos }^{4}x$

$=\int \frac{\frac{\sin x}{{\cos }^{3}x}}{\frac{{\sin }^{4}x}{{\cos }^{4}x}}dx=\int \frac{\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx$

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

$\int \frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx=\int \frac{\frac{dt}{2}}{t^2+1}=\frac{1}{2}\int \frac{dt}{t^2+1}=\frac{1}{2}{\tan }^{-1}t+C$

$=\frac{1}{2}{\left|{\tan }^{-1}\left({\tan }^{2}x\right)\right|}_0^{\pi /4}+C=\frac{1}{2}{\tan }^{-1}1$