Ashlynn Rhodes

2022-03-16

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

klepbroek31s

Beginner2022-03-17Added 4 answers

The elliptic integral which gives L is

$\frac{L}{4}=J(a,b){\textstyle \phantom{\rule{0.222em}{0ex}}}={\int}_{0}^{\frac{\pi}{2}}\sqrt{{a}^{2}{\mathrm{cos}}^{2}\theta +{b}^{2}{\mathrm{sin}}^{2}\theta},d\theta$

and it is clearly symmetric,$J(a,b)=J(b,a)$ . The change of variables $t=b\mathrm{tan}\theta$ gives

$\sqrt{{a}^{2}{\mathrm{cos}}^{2}+{b}^{2}{\mathrm{sin}}^{2}\theta}=\mathrm{cos}\theta \sqrt{{a}^{2}+{t}^{2}},\text{}\mathrm{cos}\theta =\frac{b}{\sqrt{{b}^{2}+{t}^{2}}}$

$d\theta =\frac{b}{{b}^{2}+{t}^{2}}dt$

thus

$J(a,b)={\int}_{0}^{\mathrm{\infty}}\frac{b}{{b}^{2}+{t}^{2}}dt$

thus

$J(a,b)={\int}_{0}^{\mathrm{\infty}}\frac{{b}^{2}\sqrt{{a}^{2}+{t}^{2}}}{{({b}^{2}+{t}^{2})}^{\frac{3}{2}}}dt$

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

$\frac{1}{{b}^{2}}{\int}_{0}^{\mathrm{\infty}}\frac{\sqrt{{b}^{2}+{t}^{2}}}{{({a}^{2}+{t}^{2})}^{\frac{3}{2}}}dt$

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 $\frac{L}{{\left(ab\right)}^{2}}$

and it is clearly symmetric,

thus

thus

Applying the same change of variables to 1/4 of your integral (that is, with integration from 0 to

which, from comparison with the above, clearly equals

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