Given $f(z) = \frac{e^{kz}}{z}$, which has a simple pole at z = 0 and lies inside C

$$\therefore Res(z = 0) = \lim_{z \rightarrow 0} z \times \frac{e^{kz}}{z} = 1$$

Hence $\int\frac{e^{kz}}{z} = 2\pi i ...........................(1)$

Consider,

$$I = \int\limits_{0}^{2\pi} e^{kcos\theta} \cdot e^{iksin\theta}d\theta = \int\limits_{0}^{2\pi} e^{k (cos\theta + isin\theta)}d\theta = \int\limits_{0}^{2\pi} e^{ke^{i\theta}}d\theta$$

Put $z = e^{i\theta}$ in above equation

$\therefore d\theta = \frac{dz}{iz}$ C:|z| = 1, we get

$\therefore I = \int e^{kz} \cdot \frac{dz}{iz} = \frac{1}{i} * 2\pi i = 2\pi................(from 1) \\ i.e. \int\limits_{0}^{2\pi} e^{kcso\theta} \cdot e^{iksin\theta}d\theta = 2\pi \\ i.e \int\limits_{0}^{2\pi} e^{kcos\theta} \cdot [cos(ksin\theta) + isin(ksin\theta)]d\theta = 2\pi$

Equating real and imaginary part, we get

$$\int\limits_{0}^{2\pi} e^{kcos\theta}cos(ksin\theta)d\theta = 2\pi$$

Thus $$\int\limits_{0}^{\pi} e^{kcos\theta}cos(ksin\theta)d\theta = \pi$$ hence proved