Answer
$$\int\cos^3xdx=\sin x-\frac{\sin^3x}{3}+C$$
Work Step by Step
$$A=\int\cos^3xdx$$
Following the integral form $\int\sin^mx\cos^nxdx$, this is Case 2, where $m=0$ and $n=3$.
$$A=\int\cos^2x(\cos xdx)$$ $$A=\int(1-\sin^2x)d(\sin x)$$
We set $u=\sin x$.
$$A=\int (1-u^2)du$$ $$A=u-\frac{u^3}{3}+C$$ $$A=\sin x-\frac{\sin^3x}{3}+C$$
