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