TITLE: Change of variables
QUESTION [5 upvotes]: How can I convert the integral $\int_0^{2\pi} (a^2 \cos^2 t +b^2\sin^2 t)^{-1} dt$ into an integral $\oint_\gamma z^{-1} dz$ where $z\in \mathbb C$ and $\gamma: {x^2\over a^2}+{y^2\over b^2}=1$? I can see that $|z|^2 = a^2 \cos^2 t +b^2\sin^2 t$ but the Jacobian seems very messy and I can't get the desired form.
Edit: Perhaps into a form not exactly $\oint_\gamma z^{-1} dz$ but up to a constant multiple?

REPLY [4 votes]: With
$$z=a\cos t+\mathrm ib\sin t\;,\\\bar z=a\cos t-\mathrm ib\sin t\;,$$
we have
$$\mathrm dz=(-a\sin t+\mathrm ib\cos t)\mathrm dt$$
and thus
$$\bar z\mathrm dz=\left(\mathrm i ab+(b^2-a^2)\sin t\cos t\right)\mathrm dt\;.$$
Thus
$$
\oint_\gamma z^{-1}\mathrm dz
=
\oint_\gamma\frac{\bar z\mathrm dz}{\bar zz}
=
\int_0^{2\pi}\frac{\mathrm i ab+(b^2-a^2)\sin t\cos t}{a^2\cos^2t+b^2\sin^2t}\mathrm dt\;.
$$
The first term is $\mathrm iab$ times your integral, and the second vanishes since the integrand is an odd function of $t$ and the integral is over a full period.