Answer
$$-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+c $$
Work Step by Step
Given $$\int \sin ^{5} x d x = \int (1-\cos^2x)^2\sin x d x $$
Let $$u= \cos x\ \ \to \ \ du=-\sin xdx $$
Then
\begin{aligned} \int \sin ^{5} x d x &=\int\left(1-\cos ^{2} x\right)^{2} \sin x d x \\ &=-\int\left(1-u^{2}\right)^{2} d u \\ &=-\int\left(1-2 u^{2}+u^{4}\right) d u \\ &=-\left[u-2 \frac{u^{3}}{3}+\frac{u^{5}}{5}\right]+C \\ &=-u+2 \frac{u^{3}}{3}-\frac{u^{5}}{5}+C \\ &=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+c \end{aligned}
