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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?

Américo Tavares
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    $$x=3\sin t\Longrightarrow \frac{x}{3}=\sin t\Longrightarrow t=\arcsin\left(\frac{x}{3}\right)$$ – Norbert Jun 03 '12 at 23:40
  • How does that work exactly? Why can you just swap them around like that? –  Jun 03 '12 at 23:42
  • Why not? You substituted a temporary variable, you are just changing back to be in terms of $x$. – Argon Jun 03 '12 at 23:43
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    Which implication you don't understand? – Norbert Jun 03 '12 at 23:43
  • I corrected $x = 3\sin t$ instead of $x=\sin 3t$. – Américo Tavares Jun 03 '12 at 23:43
  • I do not understand how the book plugs back in the other stuff though, it ends with something with no sin functions. –  Jun 03 '12 at 23:44
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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.

     – Robert Mastragostino Jun 03 '12 at 23:50

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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 $$\cos A = \text{adj/hyp}=\frac{\sqrt{1+x^2}}{1}=\cos (\arcsin x) $$

Argon
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  • I do not understand the last line. –  Jun 03 '12 at 23:50
  • @Jordan This is where the triangles come in. I will edit the above :). – Argon Jun 03 '12 at 23:52
  • This is making me feel sick, how do people comprehend this stuff so quickly? I have spent 4 days and I have worked through maybe 12 problems. –  Jun 03 '12 at 23:56
  • It's because you jumped in the deep end. Coming at trigonometry in the context of integrals complicates things ridiculously. You're putting the cart before the horse, so to speak. once you're used to trig problems on their own, and how to use trig identities and triangles to convert between the various trig functions, these kinds of manipulations become second nature, and this type of solution to an integral becomes a convenient insight instead of a horrible burden. – Robert Mastragostino Jun 04 '12 at 00:00
  • I have no idea what "In more general terms, if we have f(x) and t=g−1(x)⟹x=g(t) then f(g(t))=f(x)" means, and is there really no simple way to do this? We have to use a triangle and complex trig? I forgot why hypotenuse means. –  Jun 04 '12 at 01:10
  • @Jordan The triangle bit is only needed to evaluate $\cos (\arcsin x)$ (or similar). What I meant by the statement quoted is if you have a function that you want to integrate (say $f(x)$ defined in the post), you can let $x=g(t)$ where $g(t)$ is a trig substitution. Then $f(x)=f(g(t))$, which should be easier to evaluate (in my above post, it becomes simply $\cos t$). Once evaluated, solve for $x$ and replace all the instances of $t$. – Argon Jun 04 '12 at 01:23
  • I have no idea what you are saying, it is just too complex trying to imagine with f(g(t) is. –  Jun 04 '12 at 01:26
  • It is much easier to integrate $\cos t$ then $f(x)=\sqrt{1-x^2}$. However, if $x=g(t)=\sin t$, you will note the integrals are identical! $f(x)=f(g(t))$ because $x=g(t)$. – Argon Jun 04 '12 at 01:32
  • What is g(t)? I don't know where that comes in. –  Jun 04 '12 at 01:44
  • @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. – Argon Jun 04 '12 at 01:47