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...or maybe I just don't know some specific trick with trigonometric functions?

Well, anyway, here it is: $$\int{\sin^6{x}\cos^4{x}\, dx}$$

I'm bored with it, because I get 9 integrals out of 1 and the whole thing frustrates me as hell. So, is there a simpler way of integrating this or should I just make up my mind to it and integrate and integrate and integrate and integrate...untill the end comes?

By the way, here's the answer: $$ \frac{3 x}{256}-\frac{1}{512} \sin (2 x)-\frac{1}{256} \sin (4 x)+\frac{\sin (6 x)}{1024}+\frac{\sin (8 x)}{2048}-\frac{\sin (10 x)}{5120}$$

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I suggest you have a look at the "power-reducing" section on this website. It is tedious to do... – user88595 Jun 7 '14 at 15:13
@user88595 theres a fairly non-tedious way to do this using complex analysis. Check out my answer – DanZimm Jun 7 '14 at 15:16
up vote 3 down vote accepted

Hint: $$ \sin^2 x = \frac{1 - \cos 2x}{2}, \; \cos^2 x = \frac{1 + \cos 2x}{2} $$

Further you use a generalized version of these if you're comfortable with complex numbers: $$ \cos x = \frac{e^{ix} + e^{-ix}}{2}, \sin x = \frac{e^{ix} - e^{-ix}}{2i} $$ and then use the binomial theorem: $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} $$

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So, thanks a lot, I've done everything I needed with your help, It was very helpful! – Lebesgue Jun 7 '14 at 16:46

How about this?

$$\int\sin^6x\cos^4xdx=\int\sin^4x\cos^4x(\sin^2x)dx=\frac1{32}\int\sin^42x(1-\cos2x)dx=$$ $$\frac1{32}\int\sin^42xdx-\frac1{32}\int\sin^42x\cos2xdx$$

This should cut your work down quite a bit.

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To simplify things, use the fact that $(e^{ix})^n=\cos(nx)+i\sin(nx)$, comparing real and imaginary parts, you can linearize sins and cosines of any powers.

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