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?
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 $u = \sin \theta$.