$\int \frac{\sqrt{1-36 x^2}}{x^2}\,dx$

Question

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.

Answer 1

Let $6x=\sin y\implies6dx=\cos y\ dy,-\dfrac\pi2\le y\le\dfrac\pi2$

$\implies\cos y=+\sqrt{1-(6x)^2}$

$$36\int\dfrac{\sqrt{1-(6x)^2}}{(6x)^2}dx=6\int\dfrac{\cos y}{\sin^2y}\cos y\ dy =6\int\dfrac{1-\sin^2y}{\sin^2y}dy=?$$