Question

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

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

Answer 1

Answer: (C)

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

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

Dividing each term by ${\cos }^{4}x$

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

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

$=\int \frac{2tanx\cdot se{c}^{2}x}{1+{\left({\tan }^{2}x\right)}^2}dx$

Now taking ${\tan }^{2}x=t$

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

$I=\int \frac{1}{1+t^2}dt$

$I={\tan }^{-1}\left(t\right)$

$\therefore I={\tan }^{-1}\left({\tan }^{2}x\right)+c$.