How to integrate $\int \cos^2(3x)dx$ $$\int \cos^2(3x)dx$$
The answer according to my instructor is:
$${1 + \cos(6x) \over 2} + C$$
But my book says that:
$$\int \cos^2(ax)dx = {x \over 2} + {\sin(2ax) \over 4a} + C$$
I'm not really sure which one is correct. 

There are two methods you can use. Integration by parts and solving for the integral, or the half angle formula.
Remember that the half angle formula is given by $\cos^2(x) = \frac12 (1+\cos(2x))$ and also $\sin^2(x) = \frac12(1-\cos(2x))$.
Thus $$\int \cos^2(3x) dx = \frac12 \int (1+\cos(6x)) dx = \frac12 \left(x + \frac{\sin(6x)}{6}\right) + C.$$

The method of using integration by parts goes like this (I am changing $\cos(3x)$ to $\cos(x)$ for this example).
$$\int \cos^2(x) dx = \cos(x)\sin(x) + \int \sin^2(x) dx$$ we used $u=\cos(x)$ and $dv = \cos(x)dx$. Now using the pythagorean identity we have:
$$\int \cos^2(x) dx = \cos(x)\sin(x) + \int (1-\cos^2(x)) dx$$
Which leads to:
$$2 \int \cos^2(x) dx =\cos(x)\sin(x) + \int 1 dx.$$
So we have:
$$\int \cos^2(x) dx = \frac12 \left( x + \cos(x)\sin(x) \right) +C.$$
Finally using the double angle formula for $\sin(x)$ we have:
$$\int \cos^2(x) dx = \frac12 \left( x + \frac{\sin(2x)}{2} \right) +C.$$
This is the form we would find if we used the half angle formula.
