Can this integral ∫02πdθ(a2cos2θ+b2sin2θ)32 be written in the form of a elliptic integral

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Ashlynn Rhodes

2022-03-16

Can this integral

\int_{0}^{2 π} \frac{d θ}{{(a^{2} \cos^{2} θ + b^{2} \sin^{2} θ)}^{\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) = \int_{0}^{\frac{π}{2}} \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ}, d θ

and it is clearly symmetric,

J (a, b) = J (b, a)

. The change of variables

t = b \tan θ

gives

\sqrt{a^{2} \cos^{2} + b^{2} \sin^{2} θ} = \cos θ \sqrt{a^{2} + t^{2}}, \cos θ = \frac{b}{\sqrt{b^{2} + t^{2}}}

d θ = \frac{b}{b^{2} + t^{2}} d t

thus

J (a, b) = \int_{0}^{\infty} \frac{b}{b^{2} + t^{2}} d t

thus

J (a, b) = \int_{0}^{\infty} \frac{b^{2} \sqrt{a^{2} + t^{2}}}{{(b^{2} + t^{2})}^{\frac{3}{2}}} d t

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

\frac{π}{2}

instead of to

2 π

) results in

\frac{1}{b^{2}} \int_{0}^{\infty} \frac{\sqrt{b^{2} + t^{2}}}{{(a^{2} + t^{2})}^{\frac{3}{2}}} d t

which, from comparison with the above, clearly equals

J \frac{b, a}{{(a b)}^{2}}

. Using the symmetry of J, and multiplying by 4, we conclude that your integral indeed equals

\frac{L}{{(a b)}^{2}}

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