Answer
$$\int\cos^32x\sin^52xdx=-\frac{6(\cos2x)^4-8(\cos2x)^6+3(\cos2x)^8}{48}+C$$
Work Step by Step
$$A=\int\cos^32x\sin^52xdx$$ $$A=\frac{1}{2}\int\cos^32x\sin^52xd(2x)$$
We set $a=2x$.
$$A=\frac{1}{2}\int\cos^3a\sin^5ada$$
Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 1 here, where $m=5$ and $n=3$.
$$A=\frac{1}{2}\int\cos^3a\sin^4a(\sin ada)$$ $$A=-\frac{1}{2}\int\cos^3a(1-\cos^2a)^2d(\cos a)$$
We set $u=\cos a$.
$$A=-\frac{1}{2}\int u^3(1-u^2)^2du$$ $$A=-\frac{1}{2}\int u^3(1-2u^2+u^4)du$$ $$A=-\frac{1}{2}\int(u^3-2u^5+u^7)du$$ $$A=-\frac{1}{2}\Big(\frac{u^4}{4}-\frac{u^6}{3}+\frac{u^8}{8}\Big)+C$$ $$A=-\frac{1}{2}\Big(\frac{6u^4-8u^6+3u^8}{24}\Big)+C$$ $$A=-\frac{6u^4-8u^6+3u^8}{48}+C$$ $$A=-\frac{6(\cos2x)^4-8(\cos2x)^6+3(\cos2x)^8}{48}+C$$
