Question

$\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$

Answer 1

We have,

$\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$

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

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

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

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

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

$=\int \frac{2\tan x{\sec }^{2}x}{{\left({\tan }^{2}x\right)}^2+1}dx$

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

Then,

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

We know that,

$\int \frac{dx}{1+x^2}={\tan }^{-1}x$

$={\tan }^{-1}t+C$

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

Hence, this is the answer.