Answer
$$-\frac{1}{2} \cos ^{2} x+\frac{1}{2} \cos ^{4} x-\frac{1}{6} \cos ^{6} x+c$$
Work Step by Step
\begin{aligned} \int \sin ^{5} x \cos x d x &=\int \sin ^{4} x \sin x \cos x d x \\ &=\int\left(1-\cos ^{2} x\right)^{2} \sin x \cos x d x \\ &=\int\left(1-2 \cos ^{2} x+\cos ^{4} x\right) \sin x \cos x d x \\ &=\int\left(\cos x-2 \cos ^{3} x+\cos ^{5} x\right) \sin x d x \\ &=-\frac{1}{2} \cos ^{2} x+\frac{1}{2} \cos ^{4} x-\frac{1}{6} \cos ^{6} x+c \end{aligned}
