<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] \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\\\\
& = &\int \frac{(3 \tan x + 2)\sec^2x}{(2 \tan x + 3)\sec^2 x}\,dx \\\\
& = & \int \frac{(3 \tan x + 2)\sec^2 x}{(2 \tan x + 3)(1+ \tan^2 x)}\,dx\\\\
\end{eqnarray*}
Now substitute & simplify (or simplify and substitute) with your chosen $u$-substitution. You'll want to employ partial fractions, after substituting.
$u = \tan x \implies du = \sec^2 x\,dx$.
$$\int \frac{(3 u + 2)\,du}{(2u + 3)(1+u^2)} = \int \left(\dfrac{A}{2u + 3} + \frac{Bu + C}{1 + u^2}\right)\,du$$
Solving for the coefficients of $A, B, C$, we get $$A = -\frac{10}{13},\;\;B = \frac 5{13},\;\;C = \dfrac{12}{13}$$
So our integral becomes: $$\int \frac{1}{13}\left(\frac{5u + 12}{1 + u^2}-\dfrac{10}{2u + 3}\right)\,du = \dfrac{1}{13}\left(\frac 52 \int \frac {2u\,du}{1 + u^2} + 12 \int \frac{du}{1 + u^2} - 5\int \dfrac{2\,du}{2u + 3}   \right)$$
Integrating should be straightforward. Since we well be integrating in terms of $u$, once done, don't forget to "back-substitute", replacing $u$ with $\tan x$.
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