# Evaluate the following integrals:

Question:

Evaluate $\int \sin x \sin 2 x \sin 3 x d x$

Solution:

$\int \sin x \sin 2 x \sin 3 x d x$

We can write above integral as:

$=\frac{1}{2} \int(2 \sin x \sin 2 x) \sin 3 x d x-(1)$

We know that,

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

Now, considering A as $x$ and B as $2 x$ we get,

$=2 \sin x \cdot \sin 2 x=\cos (x-2 x)-\cos (x+2 x)$

$=2 \sin x \cdot \sin 2 x=\cos (-x)-\cos (3 x)$

$=2 \sin x \cdot \sin 2 x=\cos (x)-\cos (3 x)[\because \cos (-x)=\cos (x)]$

$\therefore$ integral (1) becomes,

$=\frac{1}{2} \int(\cos x-\cos 3 x) \sin 3 x d x$

$=\frac{1}{2} \int(\cos x \cdot \sin 3 x-\cos 3 x \cdot \sin 3 x) d x$

$=\frac{1}{2}\left[\int(\cos x \cdot \sin 3 x) d x-\int(\cos 3 x \cdot \sin 3 x) d x\right]$

$=\frac{1}{4}\left[\int 2(\cos x \cdot \sin 3 x) d x-\int 2(\cos 3 x \cdot \sin 3 x) d x\right]$

Cosidering $\int 2(\cos x \cdot \sin 3 x) d x$

We know,

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

Now, considering $A$ as $3 x$ and $B$ as $x$ we get,

$2 \sin 3 x \cdot \cos x=\sin (4 x)+\sin (2 x)$

$\therefore \int 2(\cos x \cdot \sin 3 x) d x=\int \sin 4 x+\sin 2 x d x \quad-(2)$

Again, Cosidering $\int 2(\cos 3 x \cdot \sin 3 x) d x$

We know,

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

Now, considering A as $3 x$ and B as $3 x$ we get,

$2 \sin 3 x \cdot \cos 3 x=\sin (6 x)+\sin (0)$

$=\sin (6 x)$

$\therefore \int 2(\cos 3 x \cdot \sin 3 x) d x=\int \sin 6 x d x \quad-(3)$

$\therefore$ integral becomes,

$=\frac{1}{4}\left[\int 2(\cos x \cdot \sin 3 x) d x-\int 2(\cos 3 x \cdot \sin 3 x) d x\right]$

$=\frac{1}{4}\left[\int(\sin 4 x+\sin 2 x) d x-\int \sin 6 x d x\right][$ From $(2)$ and $(3)]$

$=\frac{1}{4}\left[\int \sin 4 x d x+\int \sin 2 x d x-\int \sin 6 x d x\right]$

$=\frac{1}{4}\left[\frac{-\cos 4 x}{4}+\left(\frac{-\cos 2 x}{2}\right)-\left(\frac{-\cos 6 x}{6}\right)\right]+C$

$\left[\because \int \sin (a x+b) d x=-\frac{\cos (a x+b)}{a}+c\right]$

$=\frac{1}{4}\left[\frac{\cos 6 x}{6}-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}\right]+C$

$\therefore \int \sin x \sin 2 x \sin 3 x d x=\frac{1}{4}\left[\frac{\cos 6 x}{6}-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}\right]+C$