Question

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

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

Answer 1

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

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

$I=\int \frac{\sin 2x}{({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x\cdot {\cos }^{2}x}dx$

$I=\int \frac{\sin 2x}{1-\frac{1}{2}(\sin 2x{)}^2}dx$

$=2\int \frac{\sin 2x}{1+(1-{\sin }^{2}2x)}dx$

(Let $t=\cos 2x\Rightarrow dt=-2\sin 2xdx$)

$=\int \frac{-dt}{1+t^2}$
$=-{\tan }^{-1}t+C$
$=-{\tan }^{-1}\left(\cos 2x\right)+C$