Answer
$$\frac{1}{3} \cos ^{2} x \sin x+\frac{2}{3} \sin x+C$$
Work Step by Step
Given $$ \int \cos ^{3} x d x$$
Use
$$\int \cos ^{n} x d x = \frac{1}{n} \cos^{n-1} x\sin x+\frac{(n-1)}{n} \int \cos^{n-2}x dx\\
$$
Then for $n=3$
\begin{aligned}
\int \cos ^{3} x d x &=\frac{1}{3} \cos ^{2} x \sin x+\frac{2}{3} \int \cos x d x \\
&=\frac{1}{3} \cos ^{2} x \sin x+\frac{2}{3} \sin x+C
\end{aligned}
