The integral $\int \frac{\sin ^2 x \cos ^2 x d x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2…
The integral $\int \frac{\sin ^2 x \cos ^2 x d x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2}$ is equal to (where $C$ is a constant of integration).
$\frac{1}{3\left(1+\tan ^3 x\right)}+C$
$\frac{-1}{3\left(1+\tan ^3 x\right)}+C$
$\frac{-1}{1+\cot ^3 x}+C$
$\frac{1}{1+\cot ^3 x}+C$
Solution
$\begin{aligned} & \int \frac{\sin ^2 x \cdot \cos ^2 x \mathrm{~d} x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \\ & \int \frac{\tan ^2 x \cdot \sec ^6 x \mathrm{~d} x}{\left(\tan ^5 x+\tan ^2 x+\tan ^3 x+1\right)^2} \quad \text { [dividing } N^r \text { and } D^r \text { by } \cos ^{10} x \text { ] } \\ & \int \frac{\tan ^2 x\left(\tan ^2 x+1\right)^2 \sec ^2 x \mathrm{~d} x}{\left(\tan ^2 x+1\right)^2\left(\tan ^3 x+1\right)^2} \\ & \int \frac{\tan ^2 x \sec ^2 x \mathrm{~d} x}{\left(\tan ^3 x+1\right)^2} \\ & =\frac{-1}{3\left(\tan ^3 x+1\right)}+c\left[\text { Let } \tan ^3 x+1=t\right]\end{aligned}$