Taylor expansion of $\sin(x-x^2)$ I am solving a problem that involves expanding $\sin(x-x^2)$. 
Since $$\sin(x)=x-x^3/6+x^5/120-...$$ i try to substitute $x$ to $x-x^2$ to arrive at expansion of $x-x^2-x^3/6+x^4/2-x^5/2+x^6/6+...$  
Now that would be good enough, but if I look up at this expansion on WolframAlpha I see that it is supposed to look like $x-x^2-x^3/6+x^4/2-59x^5/120+x^6/8$
So does that mean that I cannot make that substitution? Where is the mistake?

It works out OK, you obviously just had a mistake in your substitution:
$$\sin(x-x^2) \sim (x-x^2) -\frac{1}{6} \left(x-x^2\right)^3 + \frac{1}{120} \left(x-x^2\right)^5 = $$
$$x - x^2 -\frac{x^3}{6} +\frac{x^4}{2} -\frac{59
   x^5}{120} +\frac{x^6}{8}...$$
