Find $\int \sin ^{5}x\,dx$.

Solution Since the exponent 5 is odd, we write $\int \sin ^{5}x\,dx=\int \sin ^{4}x \sin x\,dx$, and use the identity $\sin ^{2}x=1-\cos ^{2}x$. $$\begin{eqnarray*} \int \sin ^{5}x\,dx&=&\int \sin ^{4}x\sin x\,dx=\int (\sin ^{2}x) ^{2}\sin x\,dx=\int (1-\cos ^{2}x)^{2}\sin x\,dx\\[4pt] &=&\int (1-2\cos ^{2}x+\cos ^{4}x)\sin x\,dx \end{eqnarray*}$$

Now we use the substitution $u=\cos x$. Then $du=-\sin x\,dx$, and $$\begin{eqnarray*} \int \sin ^{5}x\,dx&=&-\int (1-2u^{2}+u^{4})~du=-u+\dfrac{2}{3}u^{3}-\dfrac{1}{5}u^{5}+C\\[3pt] &=&-\cos x+\dfrac{2}{3}\cos ^{3}x-\dfrac{1}{5}\cos ^{5}x+C \end{eqnarray*}$$