Answer
$$\int\sin2x\cos3xdx=\frac{\cos x}{2}-\frac{\cos5x}{10}+C$$
Work Step by Step
$$A=\int\sin2x\cos3xdx$$
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 $a=2$ and $b=3$:
$$A=-\frac{\cos5x}{2\times5}-\frac{\cos(-x)}{2\times(-1)}+C$$
We have $\cos(-x)=\cos x$
$$A=-\frac{\cos5x}{10}-\frac{\cos x}{(-2)}+C$$ $$A=\frac{\cos x}{2}-\frac{\cos5x}{10}+C$$
