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

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.

• Your first step is incorrect assuming you've stated your problem correctly, which I don't think you have (you have no $x$ in the numerator). – Alexis Olson Sep 15 '16 at 4:38
• The problem is that you never used $dx$ in writing your expressions. You need to write $\int \frac{\sqrt{1-36x^2}}{x^2}\,dx$ and, since $x=\frac{1}{6} \sin \theta$ then $dx=\frac{1}{6} \cos \theta\,d\theta$. – Claude Leibovici Sep 15 '16 at 6:12

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

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

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

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 $$\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=-\frac{\sqrt{1-36x^2}+6x\sin^{-1}(6x)}{x}.$$