# How do you integrate int cos^3(3x)dx?

Mar 9, 2017

$\frac{1}{4} \sin \left(3 x\right) + \frac{1}{36} \sin \left(9 x\right) + C$

#### Explanation:

This can be a tricky one, first we need to find a way to express the function to remove the power of 3.

To do this, we can try:

${\cos}^{3} \left(\theta\right) = \cos \left(\theta\right) \cdot {\cos}^{2} \left(\theta\right)$

Use ${\cos}^{2} \theta = \frac{1}{2} + \frac{1}{2} \cos 2 \theta$ to obtain:

$\cos \left(\theta\right) \left(\frac{1}{2} + \frac{1}{2} \cos 2 \theta\right)$

$= \frac{1}{2} \cos \left(\theta\right) + \frac{1}{2} \cos \left(\theta\right) \cos \left(2 \theta\right)$

Finally we can use: $\cos \left(A\right) \cos \left(B\right) = \frac{1}{2} \left\{\cos \left(A - B\right) + \cos \left(A + B\right)\right\}$ to rearrange the last term and get:

$\frac{1}{2} \cos \left(\theta\right) + \frac{1}{4} \cos \left(\theta - 2 \theta\right) + \frac{1}{4} \cos \left(\theta + 2 \theta\right)$

=$\frac{1}{2} \cos \left(\theta\right) + \frac{1}{4} \cos \left(- \theta\right) + \frac{1}{4} \cos \left(3 \theta\right)$

Of course, the cosine function has even symmetry so:

$\cos \left(- \theta\right) = \cos \left(\theta\right)$

Which will give us the exression:

$\frac{3}{4} \cos \left(\theta\right) + \frac{1}{4} \cos \left(3 \theta\right)$

So it will naturally follow that:

$\int {\cos}^{3} \left(3 x\right) \mathrm{dx} = \int \frac{3}{4} \cos \left(3 x\right) + \frac{1}{4} \cos \left(9 x\right) \mathrm{dx}$

Which easily integrates to give:

$\frac{1}{4} \sin \left(3 x\right) + \frac{1}{36} \sin \left(9 x\right) + C$