Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 23

$$-\frac{1}{6} (\cos x)^6 +c$$

Given $$\int \cos ^{5} x \sin x d x$$ Let $$ u=\cos x\ \ \ \Rightarrow \ \ du =-\sin xdx$$ Then \begin{align*} \int \cos ^{5} x \sin x d x&=\int u^{5} d u\\ &= -\frac{1}{6}u^6 +c\\ &= -\frac{1}{6} (\cos x)^6 +c \end{align*}