<s>[INST] Integral of $\cos^4(2t)\,dt$ with bounds from $0$ to $\pi$ $$\int_0^\pi\cos^4(2t)\,dt=?$$ I have attempted this problem two different ways and got two different answers that are nowhere near the correct answer. Could you please show me detailed steps on how to work this problem out. The final answer is $\frac{3\pi}{8}$. 
 [/INST] Expanding on Alex Silva's answer:
Since
$\cos^2(x) = \frac{1+\cos(2x)}{2}$,
$\begin{array}\\
\cos^4(x)
&= \frac{(1+\cos(2x))^2}{4}\\
&= \frac{1+2\cos(2x)+\frac{1+\cos(4x)}{2}}{4}\\
&= \frac{2+4\cos(2x)+1+\cos(4x)}{8}\\
&= \frac{3+4\cos(2x)+\cos(4x)}{8}\\
\end{array}
$
so
$\cos^4(2x)
=\frac{3+4\cos(4x)+\cos(8x)}{8}
$.
The integrals of the
$\cos(4x)$
and
$\cos(8x)$
terms are zero,
so the result is
$\frac{3\pi}{8}$.
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