<s>[INST] Method to integrate $\cos^4(x)$ Here my attempts for integrating $\cos^4(x)$ in few methods. 
1st method. 
$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$
$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$
$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$
$=\frac{3}{8}+\frac{1}{2}\cos2x+\frac{1}{8}\cos4x$
So, $\int\cos^4xdx=\int(\cos^2x)^2dx$
$=\int\frac{3}{8}dx+\frac{1}{2}\int\cos2xdx+\frac{1}{8}\int\cos4xdx$
$=\frac{1}{32}(12x+8\sin2x+\sin4x)+c$
2nd method.
By using reduction formula, $\cos^4xdx=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int\cos^2xdx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int(\frac{1}{2}\cos2x+\frac{1}{2})dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{8}\int\cos2x+\frac{3}{8}\int1dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{16}\sin 2x+\frac{3}{8}x$
$=\frac{1}{4}\sin x \cos^3x+\frac{3}{8}\sin x \cos x+\frac{3}{8}x$
3rd method, By using De Moivre's formula.
4th method
$\int \cos^4xdx=\int (\frac{e^{ix}+e^{-ix}}{2})^4dx$
$=\frac{1}{16}\int (e^{4ix}+4e^{2ix}+6+4e^{-2ix}+e^{-4ix})dx$
$=\frac{1}{16}\int (e^{4ix}+e^{-4ix}+4(e^{2ix}+e^{-2ix})+6)dx$
$=\frac{1}{16}(\frac{1}{4i}(e^{4ix}-e^{-4ix})+\frac{2}{i}(e^{2ix}-e^{-2ix})+6x)+c$
$\frac{1}{16}(\frac{1}{2}(\frac{e^{4ix}-e^{-4ix}}{2i})+4(\frac{e^{2ix}-e^{-2ix}}{2i})+6x)+c$
$\frac{1}{16}(\frac{1}{2}\sin 4x+4\sin 2x+6x)+c$
$\frac{1}{8}(\frac{1}{4}\sin 4x+2\sin 2x+3x)+c$
I want to ask is there more method to integrate it? Thanks.
 [/INST] I think you gave the most straightforward methods, including the sure-to-success replacement by $e^{ix}+e^{-ix}$. Basically any other method is going to be a more obfuscated form of this one.
You could also give a shot to the rational parametrization of the unit circle, by using $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2\,dt}{1+t^2}$,
so that $\int \cos^4 x dx = \int \frac {2(1-t^2)^4dt}{(1+t^2)^5}$, expanding this ugly rational fraction at the poles, and then reusing the rational parametrization to fall back on the trigonometric result.
There might also be a way to do this using Fourier expansion, but I'm not sure I want to see this.
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