Question:
Evaluate $\int \sin ^{2} x \cos ^{2} x d x$
Solution:
$y=\int\left(1-\cos ^{2} x\right) \cos ^{4} x \sin x d x$'
Let, $\cos x=t$
Differentiating both side with respect to $x$
$\frac{d t}{d x}=-\sin x \Rightarrow-d t=\sin x d x$
$y=\int-\left(1-t^{2}\right) t^{4} d t$
$y=-\int t^{4}-t^{6} d t$
Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$
$y=-\left(\frac{t^{5}}{5}-\frac{t^{7}}{7}\right)+c$
Again, put $t=\cos x$
$y=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+c$