# Integration by Trig Substution - completely stuck

I'm trying to solve this integral, but after more than an hour I can't figure it out. I've outlined my thinking below.

$$\int \dfrac{dx}{x^2\sqrt{4x^2+9}}$$

• If we let $\ a=3$ and $\ b=2$, the radical in the denominator fits the form $\ \sqrt{b^2+a^2x^2}$. This makes me think this is a trig substation problem. From the looks of the radical in the denominator $\sqrt{9+4x^2}$, this seems like a trig substitution problem.
• I make the substitution $\ x=\dfrac{3}{2}\tan\theta$ and $\ dx=\dfrac{3}{2}\sec^2\theta$. I then have: $$\int \dfrac{3}{2}\dfrac{\sec^2\theta}{\dfrac{9}{4}\tan^2\theta\sqrt{9+4(\dfrac{9}{4}\tan^2\theta)}}d\theta$$
• I pull the constants out of the integral by the constant multiple rule: $$\dfrac{12}{18} \int \dfrac{\sec^2\theta}{\tan^2\theta\sqrt{9+4(\dfrac{9}{4}\tan^2\theta)}}d\theta$$
• After simplifying the radiand, I get $\sqrt{9(1+\tan^2\theta)}$, which allows me to eliminate the radical entirely by the Pythagorean Identity (also pulling the 3 out of the denominator): $$\dfrac{2}{9} \int \dfrac{\sec^2\theta}{\tan^2\theta \sec\theta}d\theta$$
• Then I'm stuck after canceling the $\sec\theta$ in both the numerator and the denominator. $$\dfrac{2}{9} \int \dfrac{\sec\theta}{\tan^2\theta}d\theta$$

I've tried every trig identity I know to try and rewrite $\sec\theta$ and $\tan\theta$ in a way that allows me to simplify or do something and I'm just lost at this point. Can anyone please help point me in the right direction?

• I believe you may have overcorrected your constants: the integral becomes $\ \int \ \ \frac{\frac{3}{2} \sec^2 \theta \ d\theta}{(\frac{3}{2})^2 \tan^2 \theta \ \cdot \ 3 \sec \theta} \ .$ So the integrand is correct, but the multiplier on the integral should be $\ \frac{3}{2} \cdot \frac{4}{9} \cdot \frac{1}{3} = \frac{12}{54} = \frac{2}{9} \ .$ – colormegone Feb 21 '14 at 5:46
• Thank you! I missed that entirely. – aerotwelve Feb 21 '14 at 5:51
• That is all too easy to do when you have to deal with fractions in these substitutions. (Just about all of us have "been there"...) – colormegone Feb 21 '14 at 5:53

Rewrite $\sec$ and $\tan$ in terms of sines and cosines; you'll find that
$$\int \frac{\sec \theta}{\tan^2 \theta} d\theta = \int\frac{1/\cos \theta}{\sin^2 \theta/\cos^2 \theta} d\theta = \int \frac{\cos \theta}{\sin^2\theta} d\theta$$
Now consider a substitution of $u = \sin \theta$.
• @chubakueno If you make the substitution, note that $du = \cos \theta d\theta$. The integral is therefore $$\int \frac{du}{u^2}$$ – user61527 Feb 21 '14 at 5:09