Math Help - Cauchy's integral theorem

1. Cauchy's integral theorem

Use the parametrisation of the circle by $z = e^it (0\leq t\leq 2\pi)$ to show that

$\int e^{\cos t} \cos(\sin t + t )dt = 0$

2. Originally Posted by Richmond
Use the parametrisation of the circle by $z = e^{it} (0\leq t\leq 2\pi)$ to show that

$\int\kern-20pt{\color{red}{\phantom\int}_0^{\ \ 2\pi}}\!\!\! e^{\cos t} \cos(\sin t + t )dt = 0$
Presumably this is a definite integral, from 0 to 2π?

Use the fact that $\cos x$ is the real part of $e^{ix}$ to write the integral as $\text{Re}\!\!\int_0^{2\pi}\!\!\! e^{\cos t}e^{i(\sin t+t)}dt = \text{Re}\!\!\int_0^{2\pi}\!\!\! e^{\cos t+i\sin t}e^{it}dt$. Then let $z = e^{it}$ and it becomes $\text{Re}\!\oint_C ze^z\tfrac{dz}{iz}$ (C is the unit circle), so you can use Cauchy's theorem.