Answer
$$\int\sin3x\cos2xdx=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$
Work Step by Step
$$A=\int\sin3x\cos2xdx$$
Use Formula 69a, which states that
$$\int\sin ax\cos bxdx=-\frac{\cos(a+b)x}{2(a+b)}-\frac{\cos(a-b)x}{2(a-b)}+C$$
with $a=3$ and $b=2$:
$$A=-\frac{\cos5x}{2\times5}-\frac{\cos x}{2\times1}+C$$ $$A=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$
