Trying to find: $$\int{\frac{d\theta}{\sqrt{b+a\cos\theta}}}$$ (generalizing the cases for $a,b = 1$).

I've tried two approaches, none of which have gotten me far:

  1. By noticing that $\frac{d}{d\theta}\left(\sqrt{b+a\cos\theta}\right) = \frac{-a\sin\theta}{2\sqrt{b+a\cos\theta}}$, I rewrote the integral as $$\int{\frac2{-a\sin\theta}\frac{-a\sin\theta}{2\sqrt{b+a\cos\theta}}}\,d\theta$$ Then I used integration by parts ($u= -\frac2a\csc\theta, du = \frac2a\csc\theta\cot\theta \, d\theta,;dv=\frac{-a\sin\theta}{2\sqrt{b+a\cos\theta}}d\theta, v = \sqrt{b+a\cos\theta}$) to yield $$-\frac2a\csc\theta\sqrt{b+a\cos\theta} \,- \int{\frac2a\csc\theta\cot\theta \sqrt{b+a\cos\theta}}\,d\theta = -\frac{2\sqrt{b+a\cos\theta}}{a\sin\theta} \,- \int{\frac{2\cos{\theta}\sqrt{b+a\cos\theta}}{a\sin^2{}\theta}}\,d\theta$$ But I'm not sure where to go from here. u-substitution?

  2. By Weierstrass substitution, let $t=\tan{(\frac\theta{2})} \implies \cos\theta = \frac{1-t^2}{1+t^2}, d\theta = \frac{2dt}{1+t^2}$ $$\implies \int{\frac{d\theta}{\sqrt{b+a\cos\theta}}} = \int{\frac1{\sqrt{b+a\,\frac{1-t^2}{1+t^2}}}\frac{2dt}{1+t^2}}$$ $$\int\frac{2dt}{\sqrt{b(1+t^2)+a(1-t^2)}\cdot\sqrt{1+t^2}}$$ But this is very quickly becoming a mess.

Any suggestions?

  • $\begingroup$ If I recall correctly, this is one of the integrals that is not solvable in elementary functions... $\endgroup$
    – abiessu
    Apr 22, 2022 at 0:10
  • 4
    $\begingroup$ This antiderivative is given in terms of a Elliptic integral, which in general is non-elementary. So except for a handful of special cases, there's no hope of a solution using elementary methods. $\endgroup$
    – Robert Lee
    Apr 22, 2022 at 0:18
  • 3
    $\begingroup$ Does this answer your question? integrating incomplete elliptic integral of first kind $\endgroup$
    – Robert Lee
    Apr 22, 2022 at 0:22


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