How to integrate $\int\frac{1}{\sqrt{9-x^2}}dx$? I've tried to integrate $$\int\frac{1}{\sqrt{9-x^2}}dx$$ through substitution. I changed the equation to $$\int\frac{1}{\sqrt{9-x^2}}dx\equiv\int 1(9-x^2)^\frac{-1}2$$ and took $u=9-x^2$ whereby $du=-2xdx$ ($\frac{-du}{2x}=dx$) but then I got stuck because where do I go on from there? The answer is $\arcsin(\frac{1}{3}x)+c$, but how?
This is a form that we learn to recognize as related to inverse trigonometric functions. When you have an expression in your integrand that looks like $x^2\pm a^2$ or, as in this case, $a^2-x^2$ (for $a=3$), and you can't just use $u$-substitution, the strategy is to try trig substitution.

In this case, it's actually easier than that, because if you know that $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$, then you realize that this integral can be transformed into that by juggling constants and doing a $u$-substitution of $u=\frac{x}{3}$.
It's just $\arcsin\frac{x}{3}+C$ because it's the table integral.

Also, $$\left(\arcsin\frac{x}{3}\right)'=\frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot\frac{1}{3}=\frac{1}{\sqrt{9-x^2}}.$$