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

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

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- Apr 27th 2009, 07:22 PMRichmondCauchy's integral theorem
Use the parametrisation of the circle by $\displaystyle z = e^it (0\leq t\leq 2\pi) $ to show that

$\displaystyle \int e^{\cos t} \cos(\sin t + t )dt = 0$ - Apr 28th 2009, 02:30 AMOpalg
Presumably this is a definite integral, from 0 to 2π?

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