Question

Integrate:
$\int \sin 3x\cos 4xdx$

Answer 1

$\begin{array}{c}\frac{1}{2}\times \int 2\sin 3x.\cos 4xdx\left[2\sin x.\cos y=\sin \left(x+y\right)+\sin \left(x-y\right)\right]\\ \Rightarrow \frac{1}{2}\int \left(\sin 7x+\sin -x\right)dx\\ \Rightarrow \frac{1}{2}\int \sin 7xdx-\frac{1}{2}\int \sin xdx\\ \Rightarrow -\frac{1}{2}\frac{\cos 7x}{7}+\frac{1}{2}\cos x\\ \frac{1}{2}\left[\cos x-\frac{\cos 7x}{7}\right]\end{array}$