How to integrate $\int \frac{3+2\cos x}{(2+3\cos x)^2}{dx}$ via substitution? $$\int \frac{3+2 \cos x}{(2+3 \cos x)^2}{dx}$$
$$\int \frac{2}{3(2+3 \cos x)}{dx} +\int \frac{5}{3(2+3 \cos x)^2}{dx}$$
I can't think of better substitution . Please tell me what will be better substitution for it .

Let $$I = \int\frac{3+2\cos x}{(2+3\cos x)^2}dx$$
Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x$
So we get $$I = \int\frac{3\csc^2 x+2\cot x\csc x}{(2\csc x+3\cot x)^2}dx$$
Now put $(2\csc x+3\cot x) = t\;,$ Then $(2\csc x\cot x+3\csc^2 x)dx = -dt$
So $$I = -\int\frac{1}{t^2}dt = \frac{1}{t}+\mathcal{C} = \frac{\sin x}{2+3\cos x}+\mathcal{C}$$
