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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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    It was clear enough that I replied "don't" @DonAntonio.2012-06-03

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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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    I have no idea how to do this problem/2012-06-03
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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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    instead of differentiating $cot$, differentiate $-1/4cot$ instead. The point is that multiplying by a constant isn't affected by derivatives and integrals, so if you ever see a close solution that's off by a constant you can just nudge it slightly to get the answer you need. I really suggest doing less complicated examples. You need to have the basic integral properties understood like the back of your hand before you attempt more complicated questions like these.2012-06-04