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 $\frac{3\pi}{8}$. 
 A: $$I=\int_0^\pi\cos^4(2t)dt$$
Substitute: $u=2t\rightarrow du=2dt$,
$$I=\dfrac{1}{2}\int_{0\times 2}^{\pi\times 2}\cos^4(u)du=\dfrac{1}{2}\int_0^{2\pi}\cos^4(u)du$$
Apply the reduction formula: $\int\cos^n(x)dx=\dfrac{\sin(x)\cos^{n-1}(x)}{n}+\dfrac{n-1}{n}\int\cos^{n-2}(x)dx$, with $n=4$ :
$$I=\left.\dfrac{1}{8}\sin(u)\cos^3(u)\right\vert_0^{2\pi}+\dfrac{3}{8}\int_0^{2\pi}\cos^2(u)du$$
Use the trigonometric identity: $\cos^2(x)=\dfrac{1}{2}\cos(2x)+\dfrac{1}{2}$,
$$I=0+\dfrac{3}{8}\int_0^{2\pi}\left(\dfrac{1}{2}\cos(2u)+\dfrac{1}{2}\right)du$$
Substitute: $v=2u\rightarrow dv=2du$,
$$I=\dfrac{3}{32}\int_{0\times 2}^{2\pi\times 2}\cos(v)dv+\dfrac{3}{16}\int_0^{2\pi}du$$
$$I=\left.\dfrac{3}{32}\sin(v)\right\vert_0^{4\pi}+\left.\dfrac{3}{16}u\right\vert_0^{2\pi}$$
$$I=0+\dfrac{3}{16}\left(2\pi-0\right)$$
$$I=\dfrac{3\pi}{8}$$
Conclude:
$$\int_0^{\pi}\cos^4(2t)dt=\dfrac{3\pi}{8}$$
A: Hint:
Use the trigonometric identity bellow twice
$$\cos^2(x) = \frac{1+\cos(2x)}{2}.$$
A: Use this reduction formula:
\begin{align}
\int \cos^n\left(\theta\right)\:d\theta & = \frac{\cos^{n-1}\left(\theta\right)\sin\left(\theta\right)}{n}+\frac{n-1}{n}\int\cos^{n-2}\left(\theta\right)\:d\theta.\tag{1}
\end{align}
See here for an adequate derivation of the solution. There's also one for $\int\sin^{n}\left(\theta\right)\:d\theta$ and many others.
A: 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
$\frac{3\pi}{8}$.
