Is this integral with sine and cosine such a challenge? ...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:
$$ \frac{3 x}{256}-\frac{1}{512} \sin (2 x)-\frac{1}{256} \sin (4 x)+\frac{\sin (6 x)}{1024}+\frac{\sin (8 x)}{2048}-\frac{\sin (10 x)}{5120}$$

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.
