Integrate $\frac{1}{\sqrt{1 - x^2}}$ I have to calculate $\int \frac{1}{\sqrt{1 - x^2}} \operatorname{d}x$ forwards, using known rules like partial integration or substitution. 
What I'm not allowed to do is simply show that $\frac{\operatorname{d}}{\operatorname{d} x} \arcsin x = \frac{1}{\sqrt{1 - x^2}}$, but I don't see how I can use the proof backwards for integration either…
Any pointers?
 A: There is a standard substitution in this sort of situation, namely $x=\sin\theta$,
where we assume $-\pi/2 \le \theta \le \pi/2$.  Then $dx=\cos\theta\, d\theta$, and $\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=\cos\theta$ since in our interval $\cos$ is non-negative.
Thus
$$\int \frac{dx}{\sqrt{1-x^2}}=\int \frac{\cos\theta}{\cos\theta}d\theta=\int d\theta=\theta+C.$$
But $\theta=\arcsin x$. Now it's over.
Comment 1: Regrettably, it is commonplace in solutions not to mention $-\pi/2 \le \theta \le \pi/2$, and it is commonplace to not justify $\sqrt{\cos^2\theta}=\cos\theta$.  So in an integration question, in most calculus courses, the solution would be even shorter.
Comment 2: Note how close this truly standard approach is, in this case, to the suggestion by David Speyer.  The difference is that the calculus teacher would not notice.  The same substitution is used in many other integrals that involve $\sqrt{1-x^2}$, and close relatives can be used for integrals that involve $\sqrt{a-bx^2}$ where $a$ and $b$ are positive.  
A: I suspect I am going to annoy your calculus teacher by writing this but:
Suppose that you are given the problem of computing $\int f(x) dx$. A little fairy comes and whispers in your ear that the answer is $g(x)$. Then you can compute this integral in a "forward" way by making the substitution $x = g^{-1}(u)$. When you do this, $f(x) dx$ should turn into $du$, which can be integrated without difficulty.
A: Try substituting $x = cos(\theta)$.
A: Using the substition $x=\sin t$, $dx = \cos t ~dt$, we get:
$$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{\cos t ~dt}{\sqrt{1-\sin^2 t}} = \int \frac{\cos t ~dt}{\cos t} = \int dt = t.$$
By our substition $x=\sin t$  we have  $t=\arcsin x$.
Therefore
$$\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C.$$
