Question

$\int 2\sin 5x\sin 3xdx=$

• A. $\frac{\sin 2x}{4}-\frac{\sin 8x}{16}+c$
• B. $\frac{1}{8}[4\sin 2x-\sin 8x]+c$
• C. $\frac{1}{16}[4\cos 2x-\cos 8x]+c$
• D. $\frac{1}{16}[4\cos 2x+\cos 8x]+c$

Answer 1

The correct option is B $\frac{1}{8}[4\sin 2x-\sin 8x]+c$

Let $I=\int 2\sin 5x\sin 3xdx$
We know that,
$2\sin A\sin B=\cos (a-b)-\cos (a+b)$
So,
$2\sin 5x.\sin 3x=\cos \left(2x\right)-\cos \left(8x\right)$

Then,
$I=\int [\cos \left(2x\right)-\cos \left(8x\right)]dx$
$=\int \left(\cos 2x\right).dx$
$\because \int \cos 2xdx$ $=\frac{\sin 2x}{2}$
$\int \cos 8xdx$ $=\frac{\sin 8x}{8}$

$\therefore I=\frac{\sin 2x}{2}-\frac{\sin 8x}{8}+c$
$=\frac{1}{8}[4\sin 2x-\sin 8x]+c$
Hence, $\int 2\sin 5x\sin 3x=\frac{1}{8}[4\sin 2x-\sin 8x]+c$