<s>[INST] Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$ Please someone give me an idea to evaluate
this:    $$\int \frac1{(2+3\cos x)^2}\mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $\cos^2x$ does not work, so help me here. 
 [/INST] Integrating by parts,
$$\int\dfrac{dx}{(a+b\cos x)^2}=\int\dfrac{\sin x}{(a+b\cos x)^2}\cdot\dfrac1{\sin x}dx$$
$$=\dfrac1{\sin x}\int\dfrac{\sin x}{(a+b\cos x)^2}dx-\left(\dfrac{d(1/\sin x)}{dx}\cdot\int\dfrac{\sin x}{(a+b\cos x)^2}dx\right)dx$$
$$=\dfrac1{b\sin x(a+b\cos x)}+\int\dfrac{\cos x}{b(1-\cos^2x)(a+b\cos x)}\,dx$$
Use Partial Fraction Decomposition,
$$\dfrac{\cos x}{(1-\cos^2x)(a+b\cos x)}=\dfrac A{1+\cos x}+\dfrac B{1-\cos x}+\dfrac C{a+b\cos x}$$
The first two integral can be managed easily, for the last use  Weierstrass substitution
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