Answer
$$\int8\cos^32\theta\sin2\theta d\theta=2\sin^22\theta-\sin^42\theta+C=-\cos^4 2\theta+C$$
Work Step by Step
$$A=\int8\cos^32\theta\sin2\theta d\theta$$ $$A=\int4\cos^32\theta\sin2\theta d(2\theta)$$
We set $a=2\theta$.
$$A=\int4\cos^3a\sin ada$$
We would rewrite $$\cos^3a=\cos^2a\cos a=(1-\sin^2a)\cos a$$
Therefore,
$$A=\int4(1-\sin^2a)\sin a(\cos a da)$$ $$A=\int4(1-\sin^2a)\sin a d(\sin a)$$
Take $u=\sin a$
$$A=\int4(1-u^2)udu$$ $$A=4\int(u-u^3)du$$ $$A=4\Big(\frac{u^2}{2}-\frac{u^4}{4}\Big)+C$$ $$A=2u^2-u^4+C$$ $$A=2\sin^22\theta-\sin^42\theta+C$$
or (by using the trigonometric identity formula, $\sin^2+\cos^2=1$):
$$A=-\cos^4 2\theta+C$$
