Question:
Evaluate $\int \sin ^{5} x d x$
Solution:
$y=\int\left(1-\cos ^{2} x\right)^{2} \sin x d x$
Let, $\cos x=\mathrm{t}$
Differentiating both side with respect to $x$
$\frac{d t}{d x}=-\sin x \Rightarrow-d t=\sin x d x$
$y=-\int\left(1-t^{2}\right)^{2} d t$
$y=-\int 1+t^{4}-2 t^{2} d t$
Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ and $\int c d t=c t$
$y=-\left(t+\frac{t^{5}}{5}-2 \frac{t^{3}}{3}\right)+c$
Again, put $t=\cos x$
$y=-\left(\cos x+\frac{\cos ^{5} x}{5}-2 \frac{\cos ^{3} x}{3}\right)+c$