<s>[INST] 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 $J$. 
 [/INST] Note that $$3\sin x+4\cos x=5\Big( \dfrac{3}{5}\sin x+\dfrac{4}{5}\cos x \Big)$$ and $$\cos2x=1-2\sin^2x$$
also $\arcsin(4/5)=\arccos(3/5)=0.92$
combining all this, your integral becomes $\frac{1}{5}\int \frac{1 -2\sin^2x\: dx}{\sin(x+0.92)}$.
From here, writing $x=(x+0.92)-0.92$ in numerator and basic integration formula for $\csc x$, $\cot x$ will work.
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