Proving the integral $\int{1\over \sqrt{1-x^2}}\,dx$ via substitution $\int{1\over \sqrt{1-x^2}}\,dx$
How do I solve this without $\sin$ or $\cos$ substitution? I want to try using $u$ substitution because I feel like it will be more intuitive for me. So far, I have tried to make the function to the power of a negative exponent:
${(\sqrt{1-x^2})}^{-1}$
I tried to do $u$ $=$ $1-x^2$, and do the substitution method, but I keep getting the wrong answer; I'm trying to use this indefinite integral to solve a definite one from $1/2$ to $\sqrt{3}$/$2$. 
I'd rather not use trigonometric substitution because I want to do this assuming I don't already know that this is the derivative of $\sin^{-1}x$.
 A: Let $y(x)$ be defined on $(-1,1)$ as
$$y(x)=\int_0^x \frac1{\sqrt{1-t^2}}\,dt\tag1$$
Note that $y(0)=0$.

Differentiating $(1)$, we find that 
$$\frac{dy}{dx}=\frac1{\sqrt{1-x^2}}$$
whence we see that 
$$\frac{dx}{dy}=\sqrt{1-x^2}\tag2$$
Note that $x'(0)=1$.

Differentiating $(2)$ reveals
$$\begin{align}
\frac{d^2x}{dy^2}&=-\frac{x}{\sqrt{1-x^2}}\,\frac{dx}{dy}\\\\
&=-x\tag3
\end{align}$$

The general solution to the ODE of $(3)$ is $x(y)=A\sin(y)+B\cos(y)$.  Given that $x(0)=0$ and $x'(0)=1$, we find that $x(y)=\sin(y)$.  This implies that $y(x)$ is the inverse function of the sine function.  
Denoting this inverse function $y(x)=\arcsin(x)$ yields the result
$$\int_0^x \frac1{\sqrt{1-t^2}}\,dt=\arcsin(x)$$
A: Try the self similar substitution 
 $$x=\frac{1-t}{1+t},\quad t\ge 0$$
$$\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}=-\int{\frac{1}{\sqrt{t}\left( 1+t \right)}dt}=-2\int{\frac{1}{\left( 1+{{u}^{2}} \right)}du}$$
A: It is not really possible to avoid a trigonometry-less solution, as the antiderivative is $\arcsin$ and nothing else and is not expressible as a rational function (though it has a logarithmic expression in the imaginary numbers).
The nice solution by @logo is right, but strictly speaking it should also require a substitution such as $u=\tan \theta$.
In fact,
$$\frac1{\sqrt{1-x^2}}$$ should be considered an elementary integral as it appears in a table of derivatives of the common functions, just like, say, $\dfrac1x$.
A: Using the inverse function theorem, you know that the antiderivative will be a solution of the ODE
$$y'=\sqrt{1-y^2}.$$
This should ring the bell of the sine function, for which we know that $(\sin y)'=\cos y$ and $\cos y=\sqrt{1-\sin^2y}$.
Then, 
$$x=\sin y\iff y=\sin^{-1}x.$$
A: To avoid trig substitutions, apply $x=i t $ instead
\begin{align}\int{1\over \sqrt{1-x^2}}\,dx
=&\int{i\over \sqrt{1+t^2}}\,dt= i \ln\left(t +\sqrt{1+t^2}\right)\\
=&\ i \ln\left(\sqrt{1-x^2}-i x\right)
= i \ln\bigg( e^{-i\tan^{-1}\frac{x}{\sqrt{1-x^2}}}\bigg)\\
=& \ \tan^{-1}\frac{x}{\sqrt{1-x^2}}
\end{align}
