Question

Evaluate $\int {\sin x \cdot \sin 2x \cdot \sin 3xdx} $

Answer 1

Hint: Here as you can see the equation is in the form of $\sin A\sin B$, so we apply the formula and then simplify the integral.

Complete step-by-step answer:
As you know
$\sin A\sin B = \dfrac{1}{2}\left( {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right)$
Applying this, we get
$ \Rightarrow \int {\dfrac{1}{2}\left( {\cos \left( {x - 2x} \right) - \cos \left( {x + 2x} \right)} \right)\sin 3xdx} $
We know that $\cos ( - \theta ) = \cos (\theta )$
$ \Rightarrow \int {\dfrac{1}{2}\left( {\cos x - \cos 3x} \right)\sin 3xdx} $
Now break the integration
$ \Rightarrow \int {\dfrac{1}{2}\left( {\cos x\sin 3x} \right)dx - \int {\dfrac{1}{2}\left( {\cos 3x\sin 3x} \right)dx} } $
As you know,
$\cos A\sin B = \dfrac{1}{2}\left( {\sin \left( {B + A} \right) + \sin \left( {B - A} \right)} \right)$ , and $\sin A\cos A = \dfrac{1}{2}\sin 2A$
Applying this, we get
$ \Rightarrow \int {\dfrac{1}{2} \times \dfrac{1}{2}\left( {\sin \left( {3x + x} \right) + \sin \left( {3x - x} \right)} \right)dx} - \int {\dfrac{1}{2} \times \dfrac{1}{2}\left( {\sin 6x} \right)dx} $
Now again break the integrals
$ \Rightarrow \int {\dfrac{1}{4}\sin 4xdx + \int {\dfrac{1}{4}} \sin 2xdx} - \int {\dfrac{1}{4}\left( {\sin 6x} \right)dx} $
Now apply integration
As we know sin(ax) integration is $\dfrac{{ - \cos (ax)}}{a}$
$ \Rightarrow \dfrac{1}{4}\left( {\dfrac{{ - \cos 4x}}{4} + \dfrac{{ - \cos 2x}}{2} - \dfrac{{ - \cos 6x}}{6}} \right)$ + C
$ \Rightarrow \dfrac{1}{4}\left( { - \dfrac{{\cos 4x}}{4} - \dfrac{{\cos 2x}}{2} + \dfrac{{\cos 6x}}{6}} \right)$ + C
$ \Rightarrow \dfrac{1}{{48}}\left( { - 3\cos 4x - 6\cos 2x + 2\cos 6x} \right)$ + C
So, this is your required answer.

Note: In this type of question first apply the trigonometry formula, then simplify the integration you will get your answer. It can also be solved by applying integration by parts but it will be tedious.