$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to
$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to
$1 / 6$
$1 / 3$
$1 / 12$
$1 / 9$
Solution
Divide $\mathrm{Nr} \& \mathrm{Dr}$ by $\cos \mathrm{x}$
$\int_0^{\pi / 4} \frac{\tan ^2 x \sec ^2 x d x}{\left(1+\tan ^3 x\right)^2} d x$ Let $1+\tan ^3 \mathrm{x}=\mathrm{t}$
$\begin{aligned}
& \tan ^2 x \sec ^2 x d x=\frac{d t}{3} \\
& \frac{1}{3} \int_1^2 \frac{\mathrm{dt}}{\mathrm{t}^2}=\frac{1}{6}
\end{aligned}$