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. 
 A: You can use the following approach:
$$\int{\frac{dx}{(2+3\cos x)^2}}=\int{\frac{dx}{(2\cos^2{\frac{x}2}+2\sin^2{\frac{x}2}+3\cos^2{\frac{x}2}-3\sin^2{\frac{x}2})^2}}=$$
$$=\int{\frac{dx}{(5\cos^2{\frac{x}2}-\sin^2{\frac{x}2})^2}}=\int\frac{dx}{\cos^4{\frac x2}(5-\tan^2{\frac x2})^2}=\left[t=\tan\frac{x}{2}\right]=$$
$$=2\int\frac{1+t^2}{(5-t^2)^2}dt=...$$
A: 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
A: Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
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Please have a look .
A: ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$\int \frac{1}{(3\cos(x)+2)^2} dx = \int \frac{1}{\left(\frac{3(1-\tan^2(\frac{x}{2}))}{\tan^2(\frac{x}{2})+1}+2\right)^2}dx$$
now substitute  $u=\tan(\frac{x}{2})$
$$\Longrightarrow 2\int \frac{u^2+1}{(u^2-5)^2} du$$ 
now you can try the new integral.
