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

I understand I need to use $x=\frac{3}{2}\tan\theta$ trigonometric substitution

So I got to: $$\int {\frac{3}{2\cos^2\theta}\over \frac{3}{2} \tan\theta \sqrt{9+9\tan^2\theta}}=\int {\cos^2\theta\over 3\tan\theta \sec\theta}d\theta$$

how should I continue from here?

  • 5
    $\begingroup$ If you want to do trig substitution, let $2x=3\tan\theta$. $\endgroup$ – André Nicolas Jan 10 '16 at 15:25
  • $\begingroup$ Convert all functions to $\sin$ and $\cos$ $\endgroup$ – Bernard Masse Jan 10 '16 at 16:14

I'll just give a hint, solve it on your own. substitute $x=\dfrac32\tan y$ and then integrate with respect to $y$. Finally substitute $y$ with $x$. now you can reduce everything to sin and substitute sinx with a variable. you will get a cos term in the numerator. i think that should be enough


Just take $x^2$ out of square root and let $u={1/x}$ and it converts into standard ${\frac{1}{\sqrt{quadratic}}}$

$${\frac{dx}{x^2\sqrt{9x^{-2}+4}}}$$ $$u={\frac1x}$$ $$du={\frac{-1}{x^2}}$$ $${\frac{-du}{\sqrt{9u^2+4}}}$$


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