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?

$$\sin(x-x^2)=\sum_{k=0}^\infty\frac{(-1)^n}{(2n+1)!}(x-x^2)^{2n+1}=\sum_{k=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}(1-x)^{2n+1}=$$
$$=x-x^2-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{120}-\ldots=x-x^2-\frac{x^3-3x^4+3x^5-x^6}{6}+\ldots=$$
$$x-x^2-\frac{1}{6}x^3+\frac{1}{2}x^4+\ldots$$
