<s>[INST] Integrating trigonometric function problem $\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx$ \begin{eqnarray*}
\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &=&  \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\
&=& \int \frac{3\tan x +2}{2\tan x +3} dx\\
&& u = \tan x \text{ and } du = \sec^2 x \ dx 
\end{eqnarray*}
Am I going in the right direction with this one? It seems like not. 
 [/INST] Another way:
$$
\frac{3\sin x+2\cos x}{2\sin x+3\cos x} = \overbrace{\frac{12}{13} + \frac5{13}\tan\left( x - \varphi \right)}^{\text{So integrate this function.}} \text{ where }\tan\varphi = \frac 2 3.
$$
I posted a proof of this trigonometric identity (it's not very hard) in response to this question, and maybe one or the other of these two should be marked as a duplicate.
</s>