Easy way to compute $\int_{-\pi}^\pi \cos^4(x)dx$. Is there an easy way to compute $$\int_{-\pi}^\pi \cos^4(x)dx\ \ ?$$ 
What I did is using $$\cos^4(x)+1-1=1+(\cos^2(x)-1)(\cos^2(x)+1)=-\sin^2(x)(1+\cos^2(x))$$
and then using formula $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
and $$\sin^2(x)=\frac{1-\cos(2x)}{2},$$
but at the end, I have to compute $$...+\int_{-\pi}^\pi\cos^2(2x)dx.$$
Using finally that $$\cos^2(2x)=\frac{1+\cos(4x)}{2},$$
I concluded (and found $\frac{3\pi}{4}$ which is correct). But is there an other (shorter) method ? 
 A: We have that
$$ \int_{-\pi}^{\pi}e^{ni\theta}\,d\theta = 2\pi\cdot \delta(n) \tag{1}$$
and since $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, by the binomial theorem it follows that
$$ \int_{-\pi}^{\pi}\cos^4(x)\,dx = \frac{1}{16}\int_{-\pi}^{\pi}\binom{4}{2}\,d\theta = \color{red}{\frac{3\pi}{4}}.\tag{2}$$
A: $$\cos^4x=\cos^2x(1-\sin^2x)=\cos^2x-\frac14\sin^22x\implies$$
$$\int_{-\pi}^\pi\cos^4x\;dx=2\int_0^\pi\cos^4x\;dx=2\int_0^\pi\cos^2x\;dx-\frac12\int^\pi_0\sin^22x\;dx=$$
$$\left.\left(x+\cos x\sin x\right)\right|_0^\pi-\left.\frac14\frac{2x-\cos2x\sin2x}2\right|_0^\pi=\pi-\frac18(2\pi)=\frac{3\pi}4$$
A: May be this method is more difficult, but it's (in my opinion) elegant. Let $f(x)=\cos^2(x)$. As you remarked,
$$f(x)=\frac{1}{2}+\frac{1}{2}\cos(2x).$$
Therefore, it's Fourier coefficient are $a_0=1$, $a_2=\frac{1}{2}$ and $a_n=0$ if $n\neq 0,2$ and $b_n=0$ fo all $n$. Using Parseval identity, and the $2\pi-$periodicity, we get $$\int_{-\pi}^\pi (f(x))^2\mathrm d x=\int_0^{2\pi}(f(x))^2\mathrm d x=\pi\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{3\pi}{4}.$$
A: The integral has four-fold symmetry, so it is equivalent to:
$$\huge 4\int_0^\frac{\pi}{2} \cos^{4}(x) \, \text{d}x = 2 \,\text{B}\left(\frac{5}{2},\frac{1}{2}\right)$$
$$\huge = 2 \, \frac{\Gamma{(\frac{5}{2})}\Gamma{(\frac{1}{2})}}{\Gamma{(3)}} = 2 \, \frac{ \frac{3}{2} \times \frac{1}{2} \times \pi}{2} = \frac{3\pi}{4}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\int_{-\pi}^{\pi}\cos^{4}\pars{x}\,\dd x} =
2\int_{0}^{\pi}\
\overbrace{\bracks{{3 \over 8} + \half\,\cos\pars{2x} + {1 \over 8}\,\cos\pars{4x}}}
^{\ds{\cos^{4}\pars{x}}}\ \,\dd x = 2\times{3 \over 8}\times\pi=
\color{#f00}{{3 \over 4}\,\pi}
\end{align}

Note that
  $\ds{\left.\int_{0}^{\pi}\cos\pars{nx}\,\dd x
\,\right\vert_{\ n\ \in\ \mathbb{Z}\,\backslash\,\braces{0}}\;\;\;\; =\ 0}$

