Question

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

Answer 1

We have,

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

On multiply and divide by 2 and we get,

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

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

Now using formula,

$2\sin A\cos B=\sin (A+B)+\sin (A-B)$

So,

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

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

$\Rightarrow \frac{1}{2}\int \sin 7xdx-\frac{1}{2}\int \sin xdx$

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

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

Hence, this is the answer.