<s>[INST] $\int \frac{\sqrt{1-36 x^2}}{x^2}\,dx$ I need help integrating
$$\int \frac{\sqrt{1-36x^2}}{x^2} \ dx$$
using trigonometric substitution.
My first step was simplifying the integral down to $$\int \frac{\sqrt{36(\frac{1}{36}-x^2)}}{x^2} \ dx$$
and use $x=\frac{1}{6} \sin \theta$ to perform trigonometric substitution. I then perform trigonometric substitution as so
$$6\int \frac{\sqrt{\frac{1}{36}(1-\sin^2\theta)}}{\frac{1}{36}\sin^2\theta} \  d\theta$$ 
Could someone please perform the next couple steps so I can find my error? I keep getting the incorrect answer after this last step. Thank you.
 [/INST] Consider the following substitution $x=\frac{\sin (u)}{6}$ and $dx=\frac{\cos(u)}{6}$. Then, $\sqrt{1-36x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\sin^{-1}(6x)$. Hence,
$$
\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=6\int \cot^2(u) \,du=6\int (\csc^2(u)-1) \,du=-6\cot(u)-6u.
$$
Substituting back, we get
$$
-6\cot (\sin^{-1}(6x))-6\sin^{-1}(6x).
$$
Finally, using the identity $\cot(\sin^{-1}(z))=\frac{\sqrt{1-z^2}}{z}$, we get the simplified expression
$$
\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=-\frac{\sqrt{1-36x^2}+6x\sin^{-1}(6x)}{x}.
$$
</s>