Proof:Taylor expansion of inverse trigonometric functions I find it quite difficult to remember the Taylor expansion of inverse trigonometric functions.Actually in school we have been just taught the series (for finding limits in calculus without teaching us the proof).
Can someone provide a short proof of these expansions (mainly for inverse trigonometric functions) so that I can derive them even if I forget.

You can remember the generalized binomial formula 
$$(1+x)^p=1+px+\frac{p(p-1)x^2}2+\frac{p(p-1)(p-2)x^3}{3!}+\frac{p(p-1)(p-2)(p-3)x^4}{4!}\cdots,$$
and apply it for $p=-1$ (fairly easy)
$$\frac1{1+x}=1-x+x^2-x^3+x^4\cdots,$$
or $p=-1/2$,
$$\frac1{\sqrt{1+x}}=1-
\frac{x}2+
\frac{3\,x^2}{2^2\cdot 2}-
\frac{3\cdot5\,x^3}{2^3\cdot3!}+
\frac{3\cdot5\cdot7\,x^4}{2^4\cdot4!}\cdots.$$
Then for the $\arctan$, substitute $x^2$ for $x$ and integrate.
$$\int\frac{dx}{1+x^2}=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\frac{x^9}9\cdots,$$
and for the $\arcsin$ substitute $-x^2$ for $x$ and integrate
$$\int\frac{dx}{\sqrt{1-x^2}}=x+
\frac{x^3}{2\cdot3}+
\frac{3\,x^5}{2^2\cdot 2\cdot5}+
\frac{3\cdot5\,x^7}{2^3\cdot3!\cdot7}+
\frac{3\cdot5\cdot7\,x^9}{2^4\cdot4!\cdot9}\cdots.$$

For the same "price", you also get $\ln$, the natural logarithm,
$$\int\frac{dx}{1+x}=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5\cdots.$$
