Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1Use Cauchy formula to solve $$\int _0^{2\pi} \frac{dt}{a \cos t+ b \sin t +c} $$ Given $\sqrt{(a^2+b^2)}=1<c$.
I tried a variable substitution, but nothing elegant. Can anyone solve it using Cauchy formula?
Edit: there are simpler solutions, but the challenge is to solve it with complex analysis.

It may still be desirable to integrate as follows,
\begin{align}
& \int_0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c}\\
=& \int_0^{2\pi} \frac{dt}{\sqrt{a^2+b^2}\cos (t-\theta) +c}
= \int_0^{\pi} \frac{2dt}{\sqrt{a^2+b^2}\cos t +c}\\
= & \int_0^{\pi} \frac{2dt}{2\sqrt{a^2+b^2}\cos^2\frac t2+c-\sqrt{a^2+b^2}}\\
 = & \int_0^{\pi} \frac{2\sec^2\frac t2 dt}{(c+ \sqrt{a^2+b^2} )+( c- \sqrt{a^2+b^2} )\tan^2\frac t2}\\
=& \frac{4}{\sqrt{c^2-a^2-
b^2} }\tan^{-1} \left( \sqrt{\frac{c-\sqrt{a^2+b^2}}{c+\sqrt{a^2+b^2}}  }\tan\frac t2 \right)_0^{\pi}\\
=& \frac{2\pi}{\sqrt{c^2-a^2-b^2} }
\end{align}
