Maclaurin expansion of $\arcsin x$ I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that
$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$
I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!

As has been mentioned in other answers, the series for $\frac1{\sqrt{1-x^2}}$ is most easily found by substituting $x^2$ into the series for $\frac1{\sqrt{1-x}}$. But for fun we can also derive it directly by differentiation. 
To find $\frac{\mathrm d^n}{\mathrm dx^n}\frac1{\sqrt{1-x^2}}$ at $x=0$, note that any factors of $x$ in the numerator produced by differentiating the denominator must be differentiated by some later differentiation for the term to contribute at $x=0$. Thus the number of contributions is the number of ways to group the $n$ differential operators into pairs, with the first member of each pair being applied to the numerator and the second member being applied to the factor $x$ produced by the first. This number is non-zero only for even $n=2k$, and in that case it is $\frac{(2k)!}{2^kk!}$. Each such pair accumulates another factor $1\cdot3\cdot\cdots\cdot(2k-1)=\frac{(2k)!}{2^kk!}$ from the exponents in the denominator. Thus the value of the $n$-th derivative at $x=0$ is $\frac{(2k)!^2}{4^k(k!)^2}$, so the Maclaurin series is
$$
\frac1{\sqrt{1-x^2}}=\sum_{k=0}^\infty\frac1{(2k)!}\frac{(2k)!^2}{4^kk!}x^{2k}=\sum_{k=0}^\infty\frac{\binom{2k}k}{4^k}x^{2k}\;.
$$
Then integrating yields
$$
\arcsin x=\sum_{k=0}^\infty\frac{\binom{2k}k}{4^k(2k+1)}x^{2k+1}\;.
$$
