Here's how that identity is derived.

First a proof of the angle sum formula $\displaystyle \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}$

Ans 1
Now, using the angle sum formula with $\displaystyle \displaystyle \begin{align*} \alpha = \beta = X \end{align*}$ we have

$\displaystyle \displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}$

And if you're trying to use integration by parts...

$\displaystyle \displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}$

And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...