Answer
$$\frac{3\pi}{8} $$
Work Step by Step
Given
$$ \int_{0}^{\pi}\cos^4(2t)dt$$
since
$$\cos^2x=\frac{1}{2}(1+\cos 2x) $$
then
\begin{align*}
\int_{0}^{\pi}\cos^4(2t)dt&=\frac{1}{4}\int_{0}^{\pi}[1+\cos (4t)]^2dt\\
&=\frac{1}{4}\int_{0}^{\pi}[1+2\cos (4t)+\cos^2(4t)] dt\\
&=\frac{1}{4}\int_{0}^{\pi}[1+2\cos (4t)+\frac{1}{2}(1+\cos(8t))] dt\\
&=\frac{1}{4}\int_{0}^{\pi}[\frac{3}{2}+2\cos (4t)+ \cos(8t) ] dt\\
&=\frac{1}{4}\left(\frac{3}{2}t+\frac{1}{2}\sin (4t)+\frac{1}{8} \sin(8t)\right)\bigg|_{0}^{\pi}\\
&=\frac{3\pi}{8}
\end{align*}
