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I have no idea how to do this part in a problem. For example if I have $\int \frac { \sqrt{9-x^2}}{x^2}dx$

like in my book, they make $x = 3\sin t$ and then $dx = 3\cos t\, dt$

Which is fine I guess, pretty abstract but I can work through it up until the end where they get $-\cot t - t + c$

Now they want to introduce $x$ back into the function and I have no idea what to do, there is a picture of a triangle which I do not really get and then some other stuff that doesn't make sense at all (all of it). The answer involves an $\arcsin$, what do I do?

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    http://math.stackexchange.com/questions/153468/how-does-trigonometric-substitution-work/153474#comment353829_153474 is the original question, if that helps the discussion. Some of the difficulty here seems to be with the understanding of inverse trig functions themselves.2012-06-03

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Look what you stated in your post. We know

$x=3\sin t$

so

$\arcsin \left(\frac{x}{3}\right)=t$


This may make things a bit clearer (hopefully!). Say we have

$f(x)=\sqrt{1-x^2}$

We introduce a variable $t$ that suffices $x=\sin t$. In other words, you are assigning $t=\arcsin x$!

Thus $f(x)=\sqrt{1-\sin^2 t}=\cos t$

Now you have the function in terms of $x$ and $t$ - but it is simpler when in terms of $t$!

In more general terms, if we have $f(x)$ and $t=g^{-1}(x) \implies x = g(t)$ then $f(g(t))=f(x)$


You will note that if we plug $t=\arcsin x$ back in:

$\cos t=\cos (\arcsin x)=\sqrt{1-x^2}$

An easy way to see why this is is to have a triangle with angle $A=\arcsin x \implies \sin A = x$.

We know $\sin A = x = x/1 = \text{opp/hyp}$

so label the opposite side $x$ and the hypotenuse $1$. Find the third side with Pythagorean theorem, and it is $\sqrt{1^2+x^2}=\sqrt{1+x^2}$. From elementary trig, we have $\cos A = \text{adj/hyp}=\frac{\sqrt{1+x^2}}{1}=\cos (\arcsin x) $

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    @Jordan I assign $g(t)$ to just about anything I want. Choose $g(t)$ as a function that makes $f(g(t))$ simple to integrate.2012-06-04