Answer
\begin{aligned}
\int 8 \cos ^{4} 2 \pi x d x
=3 x+\frac{1}{\pi} \sin 4 \pi x+\frac{1}{8 \pi} \sin 8 \pi x+C\\
\end{aligned}
Work Step by Step
Given $$ \int 8 \cos ^{4} 2 \pi x \ \ d x $$
So, we have
\begin{aligned}
I&=
\int 8 \cos ^{4} 2 \pi x d x\\
&=
\int 8 (\cos ^{2} 2 \pi x)^2 d x\\
&\text{since} \ \ \cos ^{2} 2 \pi x=\frac{1+\cos 4 \pi x}{2} ,\text{ we get}\\
I&=8 \int\left(\frac{1+\cos 4 \pi x}{2}\right)^{2} d x\\
&=2 \int\left(1+2 \cos 4 \pi x+\cos ^{2} 4 \pi x\right) d x\\
&\text{since} \ \ \cos ^{2} 4 \pi x=\frac{1+\cos 8 \pi x}{2} ,\text{ we get}\\
I&=2 \int d x+4 \int \cos 4 \pi x d x+2 \int \frac{1+\cos 8 \pi x}{2} d x\\
&=3 \int d x+4 \int \cos 4 \pi x d x+\int \cos 8 \pi x d x\\
&=3 x+\frac{1}{\pi} \sin 4 \pi x+\frac{1}{8 \pi} \sin 8 \pi x+C\\
\end{aligned}
