Answer
\begin{aligned}
\int \cos ^{3} x \sin x d x
=-\frac{1}{4} \cos ^{4} x+C
\end{aligned}
Work Step by Step
Given $$\begin{equation}
\int \cos ^{3} x \sin x d x
\end{equation}$$
So, we have
\begin{aligned}
I&=
\int \cos ^{3} x \sin x d x\\
&=-\int \cos ^{3} x(-\sin x) d x\\
&=-\frac{1}{4} \cos ^{4} x+C
\end{aligned}
