Ashlynn Rhodes

2022-03-16

Can this integral ${\int }_{0}^{2\pi }\frac{d\theta }{{\left({a}^{2}{\mathrm{cos}}^{2}\theta +{b}^{2}{\mathrm{sin}}^{2}\theta \right)}^{\frac{3}{2}}}$ be written in the form of a elliptic integral

klepbroek31s

The elliptic integral which gives L is
$\frac{L}{4}=J\left(a,b\right)\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\frac{\pi }{2}}\sqrt{{a}^{2}{\mathrm{cos}}^{2}\theta +{b}^{2}{\mathrm{sin}}^{2}\theta },d\theta$
and it is clearly symmetric, $J\left(a,b\right)=J\left(b,a\right)$. The change of variables $t=b\mathrm{tan}\theta$ gives

$d\theta =\frac{b}{{b}^{2}+{t}^{2}}dt$
thus
$J\left(a,b\right)={\int }_{0}^{\mathrm{\infty }}\frac{b}{{b}^{2}+{t}^{2}}dt$
thus
$J\left(a,b\right)={\int }_{0}^{\mathrm{\infty }}\frac{{b}^{2}\sqrt{{a}^{2}+{t}^{2}}}{{\left({b}^{2}+{t}^{2}\right)}^{\frac{3}{2}}}dt$
Applying the same change of variables to 1/4 of your integral (that is, with integration from 0 to $\frac{\pi }{2}$ instead of to $2\pi$) results in
$\frac{1}{{b}^{2}}{\int }_{0}^{\mathrm{\infty }}\frac{\sqrt{{b}^{2}+{t}^{2}}}{{\left({a}^{2}+{t}^{2}\right)}^{\frac{3}{2}}}dt$
which, from comparison with the above, clearly equals $J\frac{b,a}{{\left(ab\right)}^{2}}$. Using the symmetry of J, and multiplying by 4, we conclude that your integral indeed equals $\frac{L}{{\left(ab\right)}^{2}}$

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