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!
 A: For $$I=\int\dfrac{dx}{a\sin^2x+b\cos^2x+c}$$
divide numerator & denominator by $\cos^2x,$
$$I=\int\dfrac{\sec^2x}{a\tan^2x+b+c(1+\tan^2x)}dx$$
set $\tan x=y$
Or divide numerator & denominator by $\sin^2x$
and then you should know :)
A: If $u=x+4$, then $du=dx$ and therefore the integral turns into:
$$\int\frac{6}{5\sin^2(u)}du$$
$$=\frac{6}{5}\int\csc^2(u)du$$
$$=\frac{6}{5}\times-\cot(u)+C$$
$$=-\frac{6\cot(x+4)}{5}+C$$
A: Hint:
By multiplying numerator and denominator by $\sec^2(x)$, you have  $$\int \frac{6\sec^2 x dx}{5\tan^2 x+4\sec^2 x}$$
Now, note that you can put everything in terms of $u=\tan x$. 
We have $6\sec^2 x dx=6 du$ and $5\tan^2 x+4\sec^2 x=5 \tan^2 x+4(\tan^2x+1)=9 \tan^2 x+4=9u^2+4.$ 
So you have $$\int \frac{6}{9u^2+4} du=(6/9)\int \frac{1}{u^2+(2/3)^2}du $$
This is a known integral involving arctan (you can also deduce it if you make the substitution $u=(2/3) \tan v$.
A: \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*}
A: Follow the steps below
$$\begin{align}
& \int \frac{6}{5\sin^{2}x+4}dx = \int \frac{6}{9-5\cos^{2}x}dx = \int \frac{6\sec^2x}{9\tan^2x+4}dx \\
=&  \int \frac{d(\frac32\tan x)}{(\frac32\tan x)^2+1} = \tan^{-1}(\frac32\tan x) +C \\
\end{align}$$
A: Considering $$I=\int\dfrac{dx}{a\sin^2(x)+b\cos^2(x)+c}$$ use the double angle to make 
$$a\sin^2(x)+b\cos^2(x)+c=\frac{1}{2} ((b-a) \cos (2 x)+a+b+2 c)$$ Now, use $x=\tan^{-1}(t)$ to make
$$I=\int \frac{dt}{ (a+c)t^2+(b+c)}=\frac 1{a+c}\int \frac{dt}{t^2+\frac{b+c}{a+c} }$$ Now, let $t=\sqrt{\frac{b+c}{a+c}}u$ to make
$$I=\frac{1}{\sqrt{(a+c) (b+c)}}\int \frac {du}{1+u^2}$$ Just finish and go back to $t$ and $x$.
