Answer
$$\frac{\sin ^{6} x}{6}-\frac{2 \sin ^{8} x}{8}+\frac{\sin ^{10} x}{10}+C $$
Work Step by Step
We integrate as follows:
\begin{align*}
\int \cos ^{5} x \sin ^{5} x d x&=\int \sin ^{5} x \cos ^{4} x \cos x d x\\
&=\int \sin ^{5} x\left(1-\sin ^{2} x\right)^{2} \cos x d x\\
&=\int \sin ^{5} x \cos x d x-2 \int \sin ^{7} x \cos x d x+\int \sin ^{9} x \cos x d x\\
&=\frac{\sin ^{6} x}{6}-\frac{2 \sin ^{8} x}{8}+\frac{\sin ^{10} x}{10}+C
\end{align*}
Where we used the fact that $\sin^2 x + \cos^2 x=1$
