<s>[INST] How do I find $ \int \frac{6}{5\sin^{2}x+4}dx $? One of the homework questions I am working on asks to show that $ \frac{\partial }{\partial x} \left(\tan^{-1}\frac{3\tan x}{2}\right)  =    \frac{6}{5\sin^{2}x+4} $ which I am able to do. However, I wonder how I should approach the reverse version of this question if it simply asks me to solve the integral
$ \int \frac{6}{5\sin^{2}x+4}dx $
My first instinct was to change $ \sin^{2}x $ into something that involves cos2x but I did not get anywhere since I had cos2x in the denominator and I did not know what to do next.
Could someone please show me how to solve this integral and more importantly how should I approach integrals in similar form like this?
Thank you so much for the help!
 [/INST] \begin{align*}
I&=\int \frac{6}{5\sin^2x+4}\,\mathrm dx\\
&=\int \frac{6}{9\sin^2x+4\cos^2x}\,\mathrm dx\\
&=\int \frac{6}{9\tan^2x+4}\frac{1}{\cos^2x}\,\mathrm dx\\
&=(let\ t = \tan(x) )\quad\int \frac{6}{9t^2+4}\,\mathrm dt\\
&=(let\ p = \frac {3}{2}t)\quad\int \frac{1}{p^2+1}\,\mathrm dp\\
&=\arctan(p)+C\\
&=(p = \frac{3}{2}\tan(x))\quad \arctan(\frac{3}{2}\tan(x))+C\\
\end{align*}
</s>