The value of $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$, is
Answer & explanation
Correct answer: option 2
We have,
$I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$
$\Rightarrow I=\int \frac{\tan ^2 x \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x$ [Dividing $N^r$ and $D^r$ by $\cos ^6 x$]
$\Rightarrow I=\frac{1}{3} \int \frac{3 \tan ^2 x \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x$
$\Rightarrow I=\frac{1}{3} \int \frac{1}{\left(1+\tan ^3 x\right)^2} d\left(1+\tan ^3 x\right)$
$=-\frac{1}{3\left(1+\tan ^3 x\right)}+C$