Find $\int \sin ^{5} x \sqrt{\cos x} dx = \int \sin^{5}x\,\cos ^{1/2}x\,dx$.

Solution The exponent of $\sin x$ is $5,$ a positive, odd integer. We factor $\sin x$ from $\sin ^{5}x$ and write $$\begin{align*} \int \sin ^{5} x\cos ^{1/2} x\,dx &= \int \sin ^{4} x\,\cos ^{1/2} x\sin x\,dx=\int(\sin ^{2}x) ^{2}\,\cos ^{1/2} x\sin x\,dx\\[4pt] &=\int ( 1-\cos ^{2}x) ^{2}\cos ^{1/2} x\sin xdx \end{align*}$$

Now we use the substitution $u=\cos x.$ $$\begin{eqnarray*} \int \sin ^{5}x\,\cos ^{1/2}x\,dx &=&\int (1-\cos ^{2}x) ^{2}\,\cos ^{1/2}x\,\sin x\,dx \nonumber \\ &\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=-\sin x~dx\)}}}{\color{#0066A7}{\hbox{\(u=\cos x\)}}}}{\color{#0066A7}{\uparrow }}}{=}&\int (1-u^{2})^{2}u^{1/2}(-du) \nonumber \\ &=&-\int (u^{1/2}-2u^{5/2}+u^{9/2})\,du=-\dfrac{2}{3}u^{3/2}+\dfrac{4}{7}u^{7/2}-\dfrac{2}{11}u^{11/2}+C \nonumber \\ &=&u^{3/2}\left[ -\dfrac{2}{3}+\dfrac{4}{7}u^{2}-\dfrac{2}{11}u^{4}\right] +C \nonumber \\ &\underset{\underset{\color{#0066A7}{\hbox{\(u=\cos x\)}}}{\color{#0066A7}{\uparrow }}}{=}&(\cos x) ^{3/2}\left[ -\dfrac{2}{3}+\dfrac{4}{7}\cos ^{2}x-\dfrac{2}{11}\cos ^{4}x\right] +C \end{eqnarray*}$$