= Solution
Parametrize the positively oriented unit <circle> by $z=e^{i\theta}$. Then $e^{-1/z}=e^{-\cos\theta+i\sin\theta}$ and $dz=ie^{i\theta}d\theta$, so
$$
\oint_Ce^{-1/z}dz
=i\int_0^{2\pi}e^{-\cos\theta}\cos(\theta+\sin\theta)d\theta
-\int_0^{2\pi}e^{-\cos\theta}\sin(\theta+\sin\theta)d\theta.
$$
Equating the <real part> and <imaginary part> with $-2\pi i$ gives
$$
\boxed{\int_0^{2\pi}e^{-\cos\theta}\cos(\theta+\sin\theta)d\theta=-2\pi,\qquad
\int_0^{2\pi}e^{-\cos\theta}\sin(\theta+\sin\theta)d\theta=0.}
$$
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