# Evaluate the following integrals:

Question:

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

Solution:

$\int \cos x \cos 2 x \cos 3 x d x$

We can write above integral as:

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

We know that,

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

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

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

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

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

$\therefore$ integral (1) becomes,

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

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

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

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

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

We know,

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

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

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

$2 \cos x \cdot \cos 3 x=\cos (4 x)+\cos (2 x)[\because \cos (-x)=\cos (x)]$

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

Cosidering $\int 2 \cos ^{2} 3 x$

We know,

$\cos 2 A=2 \cos ^{2} A-1$

$2 \cos ^{2} A=1+\cos 2 A$

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

$\int 2 \cos ^{2} 3 x=\int 1+\cos 2(3 x)=\int 1+\cos (6 x)$

$\therefore \int 2\left(\cos ^{2} 3 x\right) d x=\int 1+\cos 6 x d x \quad-(3)$

$\therefore$ integral becomes,

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

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

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

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

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

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