How does trigonometric substitution work? I have read my book, watched the MIT lecture and read Paul's Online Notes (which was pretty much worthless, no explanations just examples) and I have no idea what is going on with this at all.
I understand that if I need to find something like $$\int \frac { \sqrt{9-x^2}}{x^2}dx$$
I can't use any other method except this one. What I do not get is pretty much everything else.
It is hard to visualize the bounds of the substitution that will keep it positive but I think that is something I can just memorize from a table.
So this is similar to u substitution except that I am not using a single variable but expressing x in the form of a trig function. How does this not change the value of the problem? To me it seems like it would, algebraically how is something like $$\int \frac { \sqrt{9-x^2}}{x^2}dx$$ the same as $$\int \frac {3\cos x}{9\sin^2 x}3\cos x \, dx$$
It feels like if I were to put in numbers for $x$ that it would be a different answer.
Anyways just assuming that works I really do not understand at all what happens next.
"Returning" to the original variable to me should just mean plugging back in what you had from before the substitution but for whatever unknown and unexplained reason this is not true. Even though on problems before I could just plug back in my substitution of $u = 2x$, $\sin2u = \sin4x$ that would work fine but for whatever reason no longer works.
I am not expected to do some pretty complex trigonometric manipulation with the use of a triangle which I do not follow at all, luckily though this process is not explained at all in my book so I think I am just suppose to memorize it.
Then when it gets time for the answer there is no explanation at all but out of nowhere inverse sin comes in for some reason.
$$\frac {- \sqrt{9-x^2}}{x} - \sin^{-1} (x/3) +c$$
I have no idea happened but neither does the author apparently since there is no explanation. 

There are some basic trigonometric identities which is not hard to memorise, one of the easiest and most important being $\,\,\cos^2x+\sin^2x=1\,\,$ , also known as the Trigonometric Pythagoras Theorem.
From here we get $\,1-\sin^2x=\cos^2x\,$ , so (watch the algebra!)$$\sqrt{9-x^2}=\sqrt{9\left(1-\left(\frac{x}{3}\right)^2\right)}=3\sqrt{1-\left(\frac{x}{3}\right)^2}$$From here, the substitution proposed for the integral is $$\displaystyle{\sin\theta=\frac{x}{3}\Longrightarrow \cos\theta\, d\theta=\frac{1}{3}dx}\,\,,\,x=3\sin\theta\,,\,dx=3\cos\theta\,d\theta$$  so in the integral we get$$\int \frac{\sqrt{9-x^2}}{x^2}\,dx =\int \frac{3\sqrt{1-\left(\frac{x}{3}\right)^2}}{x^2}\to\int\frac{\rlap{/}{3}\sqrt{1-\sin^2\theta}}{\rlap{/}{9}\sin^2\theta}\,\rlap{/}{3}\cos\theta\,d\theta=$$$$\int\frac{\cos\theta\,\cos\theta}{\sin^2\theta}\,d\theta$$which is what you have in your book...:)
