Question

Evaluate: $\int \sin 3x\cos 4xdx$.

Answer 1

Consider the following question.

$I={\int }^{}\sin 3x\cos 4xdx$

$=\frac{1}{2}{\int }^{}2\sin 3x\cos 4xdx$

$=\frac{1}{2}\left({\int }^{}(\sin 7x-\sin x)\right)dx$

$=\frac{1}{2}\left({\int }^{}\sin 7x\right)-\frac{1}{2}{\int }^{}\left(\sin x\right)dx$

$=-\frac{\cos 7x}{14}+\frac{1}{2}\cos x+C$

Hence, this is the required answer.