Question

$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x}$ is equal to:

• A. $2 \tan ^{-1}\left(\tan ^{2} x\right)+C$
• B. $\tan ^{-1}\left(x \tan ^{2} x\right)+C$
• C. $\tan ^{-1}\left(\tan ^{2} x\right)+C$
• D. None of the above

Answer 1

Answer: (C)

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

Answer 2

Let $I =\int \frac{\sin 2 x }{\sin ^{4} x +\cos ^{4} x } dx$
Dividing the numerator and denominator by $\cos ^{4} x$
$I=\int \frac{2 \tan x \cdot \sec ^{2} x}{\tan ^{4} x+1} d x$
Let $\tan x = u$
$\left( \sec ^{2} x \right) dx = du$
$=\int \frac{2 u }{1+ u ^{4}} du$
Let $u ^{2}= z$
$2 u du = dz$
$I =\int \frac{ dz }{1+ z ^{2}}$
$\therefore I =\tan ^{-1} z + c$
$\therefore I =\tan ^{-1} u ^{2}+ c$
$\therefore I =\tan ^{-1}\left(\tan ^{2} x \right)+ c$