Maclaurin series of $\arctan(x)$ up to degree $4$ How can I find the Maclaurin series up to degree 4 for:
$$\arctan(x)$$
Calculating the derivatives becomes complex very quickly.
Is there a special expansion for $\arctan$ like there is for $\cos(x)$ and $\sin(x)$?

If by “bgtan” you mean the arctangent, implicitly defined by
$$
\tan\arctan x=x
$$
with values in $(-\pi/2,\pi/2)$, then the Taylor expansion at $0$ (also known as MacLaurin expansion) is the famous Gregory series:
$$
\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dotsb
$$
which converges for $x\in(-1,1]$.
Computing it by derivatives is not really difficult. Consider $f(x)=\arctan x$; then
$$
f'(x)=\frac{1}{1+x^2}
$$
and so
$$
f''(x)=-\frac{2x}{(1+x^2)^2}
$$
The third derivative is
$$
f'''(x)=-2\frac{(1+x^2)^2-4x(1+x^2)}{(1+x^2)^4}=
-2\frac{1-4x+x^2}{(1+x^2)^3}
$$
You don't need to compute the fourth derivative, because $f(x)=-f(-x)$, so all derivatives of even order are zero.
Now
$$
f(0)=0,\quad
f'(0)=1,\quad
f''(0)=0,\quad
f'''(0)=-2\quad
f''''(0)=0
$$
so
$$
\arctan x=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+
\frac{f''''(0)}{24}+o(x^4)=x-\frac{x^3}{3}+o(x^4)
$$
A different approach would be to notice that
$$
\frac{1}{1+x^2}=1-x^2+x^4+o(x^4)
$$
because of geometric series. Integrating between $0$ and $x$ gives
$$
\arctan x=\int_0^x (1-t^2+t^4+o(t^4))\,dt=
x-\frac{x^3}{3}+\frac{x^5}{5}+o(x^5)
$$
