The integral \(\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x\) is equal to

Let
\(\mathrm{I} =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \)
\( =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\sin ^3 x \cos ^2 x+\cos ^5 x \sin ^2 x+\cos ^5 x\right)^2} \mathrm{~d} x \)
\( =\int \frac{\sin ^2 x \cos ^2 x}{\left[\sin ^3 x\left(\sin ^2 x+\cos ^2 x\right)+\cos ^3 x\left(\sin ^2 x+\cos ^2 x\right)\right]^2} \mathrm{~d} x \)
\( =\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} \mathrm{~d} x \)
\( =\int \frac{\sec ^2 x \tan ^2 x}{\left(1+\tan ^3 x\right)^2} \mathrm{~d} x\)
[Dividing numerator and denominator by \(\cos ^6 x\) ]
Let \(1+\tan ^3 x=t\)
Differentiating w.r.t. \(x\), we get
\(3 \tan ^2 x \sec ^2 x \mathrm{~d} x=\mathrm{dt} \)
\( \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt} \)
\( \therefore \tan ^2 x \sec ^2 x \mathrm{~d} x=\frac{1}{3} \mathrm{dt} \)
\(\therefore \mathrm{I} =\frac{1}{3} \int \frac{1}{\mathrm{t}^2} \mathrm{dt} \)
\( =\frac{-1}{3 \mathrm{t}}+\mathrm{c} \)
\( =\frac{-1}{3\left(1+\tan ^3 x\right)}+\mathrm{c}\)