<s>[INST] 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 ? 
 [/INST] The integral has four-fold symmetry, so it is equivalent to:
$$\huge 4\int_0^\frac{\pi}{2} \cos^{4}(x) \, \text{d}x = 2 \,\text{B}\left(\frac{5}{2},\frac{1}{2}\right)$$
$$\huge = 2 \, \frac{\Gamma{(\frac{5}{2})}\Gamma{(\frac{1}{2})}}{\Gamma{(3)}} = 2 \, \frac{ \frac{3}{2} \times \frac{1}{2} \times \pi}{2} = \frac{3\pi}{4}$$
</s>