any help would be very much appreciated. The question asks to evaluate the given integral using Cauchy's formula. I plugged in the formulas for $\sin$ and $\cos$ ($\sin= \frac{1}{2i}(z-1/z)$ and $\cos= \frac12(z+1/z)$) but did not know how to proceed from there.
$$\int_0^{2\pi} \frac{dθ}{3+\sinθ+\cosθ}$$
Thanks.
[Math] Cauchy’s Theorem- Trigonometric application
complex-analysis
Best Answer
Real Method
Often, this type of integral is workable using the substitution $$ z=\tan(\theta/2)\quad\text{and}\quad\mathrm{d}\theta=\frac{\mathrm{2\,d}z}{1+z^2}\\ \sin(\theta)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(\theta)=\frac{1-z^2}{1+z^2} $$ then $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)} &=\int_{-\pi}^\pi\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}\\ &=\int_{-\infty}^\infty\frac{2\mathrm{d}z}{3(1+z^2)+2z+(1-z^2)}\\ &=\int_{-\infty}^\infty\frac{\mathrm{d}z}{z^2+z+2}\\ &=\int_{-\infty}^\infty\frac{\mathrm{d}z}{(z+1/2)^2+7/4}\\ &=\frac{2\pi}{\sqrt7} \end{align} $$
Complex Method $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)} &=\oint\frac1{3+\frac1{2i}(z-\frac1z)+\frac12(z+\frac1z)}\frac{\mathrm{d}z}{iz}\\ &=\oint\frac{2z}{6z-i(z^2-1)+(z^2+1)}\frac{\mathrm{d}z}{iz}\\ &=\oint\frac{-2i}{(1-i)z^2+6z+(1+i)}\mathrm{d}z\\ \end{align} $$ The singularities of the integrand are at $\frac{-3+\sqrt7}{2}(1+i)$ and $\frac{-3-\sqrt7}{2}(1+i)$. The second is outside the unit circle, so we only need to compute the residue at the first, which is $$ \frac{-2i}{2(1-i)z+6}=\frac{-2i}{2\sqrt7} $$ Thus, the integral is $2\pi i$ times the residue: $$ \int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}=\frac{2\pi}{\sqrt7} $$