Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$

Question

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.

Answer 1

You can use the following approach:

$$\int{\frac{dx}{(2+3\cos x)^2}}=\int{\frac{dx}{(2\cos^2{\frac{x}2}+2\sin^2{\frac{x}2}+3\cos^2{\frac{x}2}-3\sin^2{\frac{x}2})^2}}=$$ $$=\int{\frac{dx}{(5\cos^2{\frac{x}2}-\sin^2{\frac{x}2})^2}}=\int\frac{dx}{\cos^4{\frac x2}(5-\tan^2{\frac x2})^2}=\left[t=\tan\frac{x}{2}\right]=$$ $$=2\int\frac{1+t^2}{(5-t^2)^2}dt=...$$

Answer 2

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.

Answer 3

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

Answer 4

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