Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate 

$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$

My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$  and
$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$
So
$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$
So
$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$
where 
$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$
I got stuck up to integrate $J$. 

HINT: set $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
 \right)  \right) ^{2}}}
$$
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right)  \right) ^{2}}{1+ \left( 
\tan \left( x/2 \right)  \right) ^{2}}}
$$
your Integrand is given by $$-1/2\,{\frac { \left( {t}^{2}+2\,t-1 \right)  \left( {t}^{2}-2\,t-1
 \right) }{ \left( 2\,t+1 \right)  \left( t-2 \right)  \left( {t}^{2}+
1 \right) }}
$$ with $t=\tan(x/2)$
and don't Forget $$dx=\frac{2dt}{1+t^2}$$
