Question

The value of $\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$ is equal to

• A. ${\cot }^{-1}\left({\tan }^{2}x\right)+c$
• B. ${\tan }^{-1}\left({\tan }^{2}x\right)+c$
• C. ${\cot }^{-1}\left({\cot }^{2}x\right)+c$
• D. ${\tan }^{-1}\left({\cot }^{2}x\right)+c$

Answer 1

The correct option is B ${\tan }^{-1}\left({\tan }^{2}x\right)+c$

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

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

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

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

Take $u=\tan x$

$\Rightarrow du={\sec }^{2}xdx$

so, $2\int \frac{u}{u^4+1}du$

Take $v=u^2$

$dv=2udu$

so, $\int \frac{1}{v^2+1}dv$

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

$={\tan }^{-1}{u}^2+C$

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