Question

Find the integral of the function
$\sin 3x\cos 4x$

Answer 1

Consider the given integral.

$I=\int \sin 3x\cos 4xdx$

We know that

$\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$

Therefore,

$I=\frac{1}{2}\int [\sin (3x+4x)+\sin (3x-4x)]dx$

$I=\frac{1}{2}\int [\sin 7x+\sin (-x)]dx$

$I=\frac{1}{2}\int [\sin 7x-\sin x]dx$

$I=\frac{1}{2}(-\frac{\cos 7x}{7}-(-\cos x))+C$

$I=\frac{1}{2}(\cos x-\frac{\cos 7x}{7})+C$

Hence, this is the answer.