Let $ \gamma (t)= e^{it} $ where $0 \leq t \leq 2 \pi.$
Evaluate $\int_{\gamma}$ $e^{z}$ $dz$ .
Use the result to show that $\int_{0}^{2\pi} e^{\cos(t)}\cos(t+ \sin(t)) dt = 0$.
I have worked out the first contour integral to be 0 but I am unsure of how the result can help in showing the other integral.
Write down direct value of integral using parametrization:
$$\oint_{\gamma}e^{z}dz=\int_{0}^{2\pi}e^{e^{it}}ie^{it}dt=i\int_{0}^{2\pi}e^{e^{it}+it}dt=\\=i\int_{0}^{2\pi}e^{\cos t+i\sin(t)+it}dt$$
But:
$$e^{i(t+\sin t)}=\cos(t+\sin(t))+i\sin(t+\sin(t))$$
So:
$$i\int_{0}^{2\pi}e^{\cos t+i\sin(t)+it}dt=i\int_{0}^{2\pi}e^{\cos(t)}\cos(t+\sin(t))dt-\int_{0}^{2\pi}e^{\cos(t)}\sin(t+\sin(t))dt$$
So $0=\Im(\oint_{\gamma}e^{z}dz)=\int_{0}^{2\pi}e^{\cos(t)}\cos(t+\sin(t))dt$.