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I need some help evaluating the integral $$ \int \frac{1}{(a+b \cos x)^{2}} \, \mathrm dx .$$ I have achieved a result by using the substitution of letting $ \tan(x/2) = t $, but it is quite cumbersome and big afterwards. Please suggest some shorter methods if possible. Any help would be appreciated.

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  • $\begingroup$ Maple also uses the tan-half angle substitution $\endgroup$ – Dr. Sonnhard Graubner Apr 9 '19 at 10:52
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Without loss of generality, $a>b>0$ $$ I(a)=\int\frac{1}{a+b\cos x}dx $$ and then $$ I'(a)=-\int\frac{1}{(a+b\cos x)^2}dx. $$ Using $u=\tan(\frac{x}{2})$, one has \begin{eqnarray} I(a)&=&\int\frac{1}{a+b\cos x}dx\\ &=&\int\frac{2}{(a-b)u^2+(a+b)}du\\ &=&\frac{2}{\sqrt{a^2-b^2}}\arctan(\sqrt{\frac{a-b}{a+b}}u)\\ &=&\frac{2}{\sqrt{a^2-b^2}}\arctan(\sqrt{\frac{a-b}{a+b}}\tan(\frac{x}{2})). \end{eqnarray} Hence $$ \int\frac{1}{(a+b\cos x)^2}dx=-I'(a)=... $$ It is straight forward to get $I'(a)$ and omit the detail. For other cases of $a$ and $b$, you can use the same way to discuss.

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Hint:

WLOG, $b=1$ for simplicity. Let $z=e^{ix}$, so that $dx=\dfrac{dz}{iz}$.

$$\int\frac{dz}{4iz(4a+z+z^{-1})^2}=\frac1{4i}\int\frac{z\,dz}{(z^2+4az+1)^2}=\frac1{4i}\int\frac{(w-2a)\,dz}{(w^2+1-a^2)^2}.$$

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