# How do I evaluate the indefinite integral intsin(3x)*sin(6x)dx ?

Aug 13, 2014

It can have two solutions

$= \frac{\sin 3 x}{6} - \frac{\sin 9 x}{18} + c$, where $c$ is a constant OR

$= \frac{2}{9} {\sin}^{3} \left(3 x\right) + c$, where $c$ is a constant

Explanation

$= \int \sin \left(3 x\right) \cdot \sin \left(6 x\right) \mathrm{dx}$

From trigonometric identities,

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

Similarly, for the given problem,

$\sin \left(3 x\right) \cdot \sin \left(6 x\right) = \frac{1}{2} \left(\cos \left(- 3 x\right) - \cos 9 x\right)$

As, $\cos \left(- A\right) = \cos \left(A\right)$

$\sin \left(3 x\right) \cdot \sin \left(6 x\right) = \frac{1}{2} \left(\cos 3 x - \cos 9 x\right)$

integrating both sides,

$\int \sin \left(3 x\right) \cdot \sin \left(6 x\right) \mathrm{dx} = \int \frac{1}{2} \left(\cos 3 x - \cos 9 x\right) \mathrm{dx}$

$= \frac{1}{2} \int \left(\cos 3 x\right) \mathrm{dx} - \frac{1}{2} \int \left(\cos 9 x\right) \mathrm{dx}$

$= \frac{1}{2} \frac{\sin 3 x}{3} - \frac{1}{2} \frac{\sin 9 x}{9} + c$, where $c$ is a constant

$= \frac{\sin 3 x}{6} - \frac{\sin 9 x}{18} + c$, where $c$ is a constant

Another Method :

$= \int \sin \left(3 x\right) \cdot \sin \left(6 x\right) \mathrm{dx}$

$= \int \sin \left(3 x\right) \cdot 2 \sin \left(3 x\right) \cos \left(3 x\right) \mathrm{dx}$

$= \int 2 {\sin}^{2} \left(3 x\right) \cos \left(3 x\right) \mathrm{dx}$

let's assume $\sin \left(3 x\right) = t$, then $3 \cos \left(3 x\right) \mathrm{dx} = \mathrm{dt}$

therefore,

$= 2 \int {t}^{2} / 3 \cdot \mathrm{dt}$

$= \frac{2}{9} \cdot {t}^{3} + c$, where $c$ is a constant

$= \frac{2}{9} {\sin}^{3} \left(3 x\right) + c$, where $c$ is a constant