Showing a bit more to the end result, we would see the following:
So we are trying to integrate the following expression $~~~\rightarrow ~~~ \dfrac{96}{6} \displaystyle\int \cos^{4} (16x)\ dx$.
To do the this, we will need to make an appropriate substitution inside of the integrand. Doing this leads us to the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{96}{6}\displaystyle\int \cos^{4} (16x)\ dx$
Let: $~u =16x$
$du=16\ dx$
$dx=\dfrac{1}{16}\ du$
Substituting in u and dx we see that we get the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\dfrac{96}{6}\cdot \dfrac{1}{16}\displaystyle\int \cos^{4} (u)\ du$
Using the reduction formula for cosine to the m power, where $m \in \mathbb{N}$. $\int \cos^{m}(u) dx = \dfrac{1}{m} \cos^{m-1}(u) \sin (u) + \dfrac{m-1}{m} \int \cos^{m-2}(u)\ dx,~ \text{where }~ m = 4,~\text{gives}:$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{96}{6}\cdot \dfrac{1}{16} \Bigg[\dfrac{1}{4} \cos^{3}(u) \sin (u) + \dfrac{3}{4} \displaystyle\int \cos^{2} (u)\ dx \Bigg]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4} \displaystyle\int \cos^{2} (u)\ dx $
Now we can use the the trigonometric identity for $\cos^{2}(u)$ and re-write it as the folllowing: $\cos^{2}(u)=\dfrac{1}{2}+\dfrac{1}{2}\cos (2u)$ . Now with this, let's replace the integrand with this identity and $u$ substitution as so,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4}\displaystyle\int \dfrac{1}{2}+\dfrac{1}{2}\cos (2u)\ du$
Which now we can integrate each separately as they are being added as a sum and also pull out any constants from the integrand as the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{4} \Bigg[ \dfrac{1}{2} \displaystyle\int \! \ du + \dfrac{1}{2} \int \cos (2u)\ du \Bigg]$
Now making another substitution, we see the following:
Let: $~w =2u$
$dw=2\ du$
$du=\dfrac{1}{2}\ dw$
Making the substitutions in for w and du we see that we get the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \displaystyle\int \ du + \dfrac{1}{2}\cdot \dfrac{1}{2}\cdot \dfrac{3}{4} \int \cos (w)\ dw$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \sin (w) + K$
Plugging back in for what $w$ is gives:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \sin (2u) + K$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{16} \cdot 2\sin (u)\cos (u) + K$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} u + \dfrac{3}{8} \sin (u)\cos (u) + K$
Plugging back in for what $u$ is gives:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \cdot (16x) + \dfrac{3}{8} \sin (16x) \cos (16x) + K$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + \dfrac{3}{8} \cdot (16x) + \dfrac{3}{8}\cdot \dfrac{1}{2} \sin (32x) + K$
Which is, $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~\dfrac{1}{4} \cos^{3} (16x) \sin (16x) + 6x + \dfrac{3}{16} \sin (32x) + K.~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$
Which can be cleaned up a bit further to this:
$\dfrac{1}{4}\Bigg[\cos^{3}(16x) \sin (16x) + \dfrac{3}{4}\sin (32x) + 24x\Bigg] + K.$
NOTE: This expression can reduced further using more identities, but not necessary. I will leave it as this stage. Just wanted to point that out in case you see or get a different solution from this here.
Okay, I hope that this has helped out. Let me know if there is any step covered that did not make much sense for doing so.
Thanks.
Good Luck.