Answer
$$\frac{1}{4} \cos^4(2- x) +c$$
Work Step by Step
Given
$$\int \cos ^{3} (2-x) \sin (2-x) d x$$
Let $$ u=\cos(2-x)\ \ \ \Rightarrow \ \ du =\sin (2-x)dx$$
Then
\begin{align*}
\int \cos ^{3} (2-x) \sin (2-x) d x&=\int u^{3} d u\\
&= \frac{1}{4}u^4 +c\\
&= \frac{1}{4} \cos^4(2- x) +c
\end{align*}
