Integral of $\int \frac {dx}{x^2 \sqrt{4-x^2}}$ 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 $t$.

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 $\int\frac{1}{4}\csc^2 t\,dt$.
