I would like to find the Maclaurin's series for $\tan(x+x^2)$ stopped at $x_0^3$.
I tried with: $$\tan(x+x^2)=1+\frac{1+2x_0}{\cos^2(x_0+x_0^2)}x+....,$$ but it seems like the derivation gets more complicated. I was wondering if there was a simpler way to solve the Maclaurin series. I was thinking about putting $t=x+x^2$, but it looks like wrong (why?).
Update:
I know that MacLaurin's expansion for tan(x): $$T(x) = x + \frac {1}{3}x^{3} + \frac { 2 } { 15 } x ^ { 5 } + \frac { 17 } { 315 } x ^ { 7 } + \ldots$$ But according to Mathematica, there should be a $x^2$ for the MacLaurin series for $tan(x+x^2)$, which I couldn't understand.
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$ $$T(x+x^2)=(x+x^2)+\frac13(x+x^2)^3+\cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$