I have arrived at definite integral with trigonometric functions

$I(a, b) = \int_0^{2 \pi} \frac{1 - a \sin(\theta) - b \cos(\theta)}{(1 +a^2 +b^2 - 2 a \sin(\theta) - 2 b \cos(\theta) )^{3/2}} \mathrm{d} \theta, \quad 0 \le a, b < 1$

I was expecting result with elliptic integrals, but Mathematica could not evaluate it :-(

Any suggestions to evaluate such integral analytically?

  • $\begingroup$ Mathematica 10.4 gives indeed a result in terms of elliptic functions. $\endgroup$ – mickep Apr 8 '16 at 3:42
  • 2
    $\begingroup$ Notice that $I(a,b)=I(\sqrt{a^2+b^2},0)=I(0,\sqrt{a^2+b^2})$. $\endgroup$ – Yves Daoust Apr 8 '16 at 8:41

Using a CAS, $$\alpha\,I(a,b)=\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right) K(z(a,b))+\left(1-a^2-b^2\right) E(z(a,b))$$ where $$z(a,b)=\frac{4 \sqrt{a^2+b^2}}{a^2+b^2+1+2 \sqrt{a^2+b^2}}$$ $$\alpha=\frac 12 \sqrt{a^2+b^2+1+2 \sqrt{a^2+b^2}}\,\left(a^2+b^2+1-2 \sqrt{a^2+b^2}\right)\,$$


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