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

Hint:

With $z=e^{it}$, $$intfrac{dt}{acos t+bsin t+c}=ointfrac{2,dz}{iz(a(z+z^{-1})-ib(z-z^{-1})+2c)}$$ and you have two nice poles.

It may still be desirable to integrate as follows,

begin{align} & int_0^{2pi} frac{dt}{acos t+ bsin t +c}\ =& int_0^{2pi} 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}{2sqrt{a^2+b^2}cos^2frac t2+c-sqrt{a^2+b^2}}\ = & int_0^{pi} frac{2sec^2frac t2 dt}{(c+ sqrt{a^2+b^2} )+( c- sqrt{a^2+b^2} )tan^2frac 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}} }tanfrac t2 right)_0^{pi}\ =& frac{2pi}{sqrt{c^2-a^2-b^2} } end{align}

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