<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] You can use the following approach:
$$\int{\frac{dx}{(2+3\cos x)^2}}=\int{\frac{dx}{(2\cos^2{\frac{x}2}+2\sin^2{\frac{x}2}+3\cos^2{\frac{x}2}-3\sin^2{\frac{x}2})^2}}=$$
$$=\int{\frac{dx}{(5\cos^2{\frac{x}2}-\sin^2{\frac{x}2})^2}}=\int\frac{dx}{\cos^4{\frac x2}(5-\tan^2{\frac x2})^2}=\left[t=\tan\frac{x}{2}\right]=$$
$$=2\int\frac{1+t^2}{(5-t^2)^2}dt=...$$
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