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For which n,m $\int_{0}^{2\pi} \cos^n(x)\sin^m (x)?$ And what are others tricks to solve these integrals in a fast way?

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    I think the Beta Function is something you might want to look at. The question reminded me of that. – 2017-02-06
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    The integral is zero whenever $m$ or $n$ is odd (shift it to the origin to integrate over symmetric integral and check that integrand is odd, you might need to do this twice). Not sure how to solve even $m$ and $n$ case fast, but it seems that it is equal to $$\frac{\pi(m-1)!!(n-1)!!}{(m+n-1)!!2^{m+n-1}}\binom{m+n}{\frac{m+n}{2}}$$ :) – 2017-02-06

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First if they're both positive integers then you can attempt to u-sub one of them out for example $\int_{0}^{2\pi} \cos^2 x\sin^3 x\text{d}x$ $\overbrace{=}^{u=\cos x} -\int_{1}^{1} u^2(1-u^2)\text{d}u$ and then proceed from there. However if we have a fractional exponent such as $\frac{1}{2}$ then the integral is undefined. Another trick you could use is trig identities $(\sin x \cos x=\frac{\sin 2x}{2})$ and power reduction for the following integral $\int_{0}^{2\pi} \sin^2 x\cos^2 x\text{d}x=\int_{0}^{2\pi}\frac{\sin^2 2x}{4}\text{d}x=$ $\int_{0}^{2\pi}\frac{1-\cos 4x}{8}\text{d}x$ and proceed from there. The final way I can think of is when dealing with something like $\int \cos ^3 x\text{d}x$ you reduce it using the Pythagorean identity $\int \cos x(1-\sin^2 x)\text{d}x$ and proceed with a u-sub, hope this helps

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The "trick" you are looking for is in fact a great "method".

As remarked by @Vivek Kaushik, it is a "Beta" integral, with the main result (See (https://en.wikipedia.org/wiki/Beta_function).):

$${\displaystyle \mathrm {B} (x,y):=2\int _{0}^{\frac {\pi }{2}}(\sin \theta )^{2x-1}(\cos \theta )^{2y-1}={\frac {\Gamma (x)\,\Gamma (y)}{\Gamma (x+y)}}.}$$

valid for any $x$ and $y$, and values of function $\Gamma$ (Gamma) that are available on every computing system (even on some pocket computers). One must know that function $\Gamma$ is an extension to any real value of function "factorial" with $\Gamma(x)=(x-1)!$. See (https://en.wikipedia.org/wiki/Gamma_function).

Of course, if the interval of integration is $(0,2\pi)$, you have to split into four parts and come back, through parity and periodicity considerations, to interval $(0,\pi/2).$