Answer
$$\int\sin2x\cos4xdx=\frac{\cos2x}{4}-\frac{\cos6x}{12}+C$$
Work Step by Step
$$A=\int\sin2x\cos4xdx$$
Recall the identity $$\sin a\cos b=\frac{\sin(a+b)+\sin(a-b)}{2}$$
Therefore, $$A=\int\frac{\sin6x+\sin(-2x)}{2}$$
Since $\sin(-a)=-\sin a$, $\sin(-2x)=-\sin2x$
$$A=\int\frac{\sin6x-\sin2x}{2}$$ $$A=\frac{1}{2}(\int \sin6x-\int\sin2x)$$ $$A=\frac{1}{2}\Big(-\frac{\cos6x}{6}+\frac{\cos2x}{2}\Big)+C$$ $$A=\frac{\cos2x}{4}-\frac{\cos6x}{12}+C$$
