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