Answer
$$\int\sin^3x\cos^3xdx=\frac{\sin^4x}{4}-\frac{\sin^6x}{6}+C$$
Work Step by Step
$$A=\int\sin^3x\cos^3xdx$$
Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 1 here, where $m=3$ and $n=3$.
$$A=\int\sin^3x\cos^2x(\cos xdx)$$ $$A=\int\sin^3x(1-\sin^2x)d(\sin x)$$
We set $u=\sin x$.
$$A=\int u^3(1-u^2)du$$ $$A=\int(u^3-u^5)du$$ $$A=\frac{u^4}{4}-\frac{u^6}{6}+C$$ $$A=\frac{\sin^4x}{4}-\frac{\sin^6x}{6}+C$$
