Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1<c$

Question

Use 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.

Answer 1

This can be done without taking the help of integration limits too.

Evaluate $$\int \frac{dt}{a \cos t+ b \sin t +c} $$ ,where $\sqrt{(a^2+b^2)}=1<c$.

I don't know what Cauchy's formula you are talking about. But here is a nice way to do it.

Since $\sin t=\frac{2\tan(\frac x2)}{1+\tan^2(\frac x2)},\cos t=\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}$, we can write it as(You may stop and try it yourself at this point) \begin{align*} &\Rightarrow\int \frac{\sec^2(\frac x2)dt}{a(1-\tan^2(\frac x2))+2b\tan(\frac x2)+c(1+\tan^2(\frac x2))}\\ &=\int\frac{2du}{a(1-u^2)+2bu+c(1+u^2)}\\ &=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+\frac{c+a}{c-a}-\frac{b^2}{(c-a)^2}}\\ &=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+\frac{c^2-(a^2+b^2)}{(c-a)^2}}\\ &=\frac2{c-a}\int\frac{du}{\left(u+\frac{b}{c-a}\right)^2+K^2}\\ &=\frac2{(c-a)K^2}\int\frac{du}{\left(\frac{u+\frac{b}{c-a}}{K}\right)^2+1}\\ \end{align*} Can you make the substitution and finish it?

Answer 2

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}

Answer 3

Hint:

With $z=e^{it}$, $$\int\frac{dt}{a\cos t+b\sin t+c}=\oint\frac{2\,dz}{iz(a(z+z^{-1})-ib(z-z^{-1})+2c)}$$ and you have two nice poles.