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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?

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    $\begingroup$ 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} \ . $ $\endgroup$ Commented Feb 21, 2014 at 5:46
  • $\begingroup$ Thank you! I missed that entirely. $\endgroup$
    – aerotwelve
    Commented Feb 21, 2014 at 5:51
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    $\begingroup$ 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"...) $\endgroup$ Commented Feb 21, 2014 at 5:53

1 Answer 1

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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 $u = \sin \theta$.

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  • $\begingroup$ I am totally new to integration, may you tell me how to continue? If that substitution is made, then the numerator gets irrational again, so I am stuck there... $\endgroup$
    – chubakueno
    Commented Feb 21, 2014 at 5:09
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    $\begingroup$ @chubakueno If you make the substitution, note that $du = \cos \theta d\theta$. The integral is therefore $$\int \frac{du}{u^2}$$ $\endgroup$
    – user61527
    Commented Feb 21, 2014 at 5:09
  • $\begingroup$ Thank you very much, understood now. $\endgroup$
    – chubakueno
    Commented Feb 21, 2014 at 5:16

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