Question

If the value of $\int (\sin 3x\cdot \cos 3x\cdot \cos 6x\cdot \cos 12x)dx=\frac{\cos ax}{b}+C,$ then the absolute value of $\frac{b}{a}$ is
(where $C$ is constant of integration)

Answer 1

$I=\int (\sin 3x\cdot \cos 3x\cdot \cos 6x\cdot \cos 12x)dx$
$\Rightarrow I=\frac{1}{2}\int \left(2\sin 3x\cos 3x\right)\cdot \cos 6x\cdot \cos 12xdx$
$\Rightarrow I=\frac{1}{4}\int \left(\right(2\cdot \sin 6x\cdot \cos 6x)\cdot \cos 12x)dx$
$\Rightarrow I=\frac{1}{8}\int \left(2\sin 12x\cos 12x\right)dx$
$\Rightarrow I=\frac{1}{8}\int \left(\sin 24x\right)dx=-\frac{\cos 24x}{8\times 24}+C$
$\therefore \frac{\cos ax}{b}=-\frac{\cos 24x}{8\times 24}$
$\Rightarrow \frac{b}{a}=-8$
$\therefore \left|\frac{b}{a}\right|=8$