Question

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

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

Answer 1

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

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

Divide $N^{\prime }and{D}^{\prime }$ by ${\cos }^{4}x.$

$\therefore I=\int \frac{2\tan x{\sec }^{2}xdx}{1+{\tan }^{4}x}$

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

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