Question

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

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

Answer 1

Answer: (A)

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

Given, $\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$\Rightarrow$ $\int \frac{2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$
Now dividing numerator and denominator by ${\cos }^{4}x$ , we have
$\int \frac{2\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx$
Now put ${\tan }^{2}x=t$ $\Rightarrow$ $2\tan x{\sec }^{2}xdx=dt$
$\Rightarrow$ $\int \frac{1}{t^2+1}dt$
$\Rightarrow$ ${\tan }^{-1}t+c={\tan }^{-1}\left({\tan }^{2}x\right)+c$ is the answer.