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I am trying to find $$\int \frac {dx}{x^2 \sqrt{4-x^2}}$$

I make $t=2\sin x$

$$\int \frac {dx}{x^2 \sqrt{4-4\sin^2 t}}$$

$$\int \frac {dx}{x^2 \sqrt{4(1-\sin^2 t)}}$$

$$\int \frac {dx}{x^2 \sqrt{4(\cos^2 t)}}$$

$$\int \frac {dx}{x^2 \cdot 2\cos t}$$

I do not really know where to go from here. I have two variables and that is really bad but I do not know how to write $x$ in terms of $t$.

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I'm afraid to say @Jordan that it already looks as if you only want someone here to solve your homework, without you trying/succeeding to get some benefit of this all (and I believe you when you say you don't like this but you only have to do it). I think that either you try really harder or else try to get some personalised help at your college/H.S. from your instructor. I, for one, can't continue to answer your questions until I see some real progress in you. – DonAntonio Jun 3 '12 at 22:43
@DonAntonio: He's certainly making progress. He added his solution to the questions today to find out what he has done wrong. If you can't continue to answer his questions, just don't answer his questions. – Gigili Jun 3 '12 at 22:51
Yes @Gigili, that's what I said: I can't. Wasn't this clear enough to you? – DonAntonio Jun 3 '12 at 23:50
It was clear enough that I replied "don't" @DonAntonio. – Gigili Jun 3 '12 at 23:52
up vote 3 down vote accepted

On many occasions if you see a factor of $x^2$ in the denominator of the integrand, you may find it convenient to get rid of it by substituting: $$x=\frac{1}{t}$$ If you are computing a definite integral, take care if the origin is on the integration interval. Hence $$dx=-\frac{dt}{t^2}$$ $$I=-\int\frac{t^{2}t \, dt}{t^{2}\sqrt{4t^{2}-1}}=-\frac{1}{8}\int\frac{d\left(4t^{2}-1\right)}{\sqrt{4t^{2}-1}}=-\frac{1}{8}2\sqrt{4t^{2}-1}=-\frac{\sqrt{4-x^{2}}}{x}$$ I have used the fact that $dx=\frac{1}{a} \, d(ax+b)$, ($a\ne0$) and $(\sqrt{t})'=\frac{1}{2\sqrt{t}}$

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What do you mean by "if the origin is on the integration interval"? – user138246 Jun 3 '12 at 23:05
if you were evaluating a definite integral and your integration limits were $a<0<b$ then $\frac{1}{t}$ has a discontinuity inside the interval and the answer would be wrong – Valentin Jun 3 '12 at 23:07
How do I know that? – user138246 Jun 3 '12 at 23:08
see the theorem i mentioned here – Valentin Jun 3 '12 at 23:15
I have no idea what that means, is that just the subsitution formula? – user138246 Jun 3 '12 at 23:17

You must replace every occurrence of $x$. Let $x=2\sin t$. This is so that the square root of $4-x^2$ will be nice. Then $dx=2\cos t\,dt$, and $\sqrt{4-x^2}=2\cos t$. So we end up needing to find $$\int \frac{2\cos t}{(4\sin^2 t)(2\cos t)}dt.$$ There is cancellation, and we end up needing to find $\int\frac{1}{4}\csc^2 t\,dt$.

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I do not see what you did to solve for x. – user138246 Jun 3 '12 at 22:49
We said let $2 \sin t$. So there is no need to solve for $x$. – André Nicolas Jun 3 '12 at 22:54
Oh I get it, I have no idea how I couldn't figure that out on my own. Also I have no idea how to get csc, I think I need to turn it back into sin but I am not sure. – user138246 Jun 3 '12 at 22:55
I have no idea what to do next, my book gets a really weird answer. Do I plug back in t or do I find the anti derivative? – user138246 Jun 3 '12 at 23:14
@Jordan, integration is the reverse of differentiation, and $cot$ is so close! Can you think of what to multiply $cot$ by, so that when differentiated you get $1/4\csc^2$? Because that (plus the constant) will be your answer. – Robert Mastragostino Jun 4 '12 at 0:29

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