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