Consider,

$$I = \int\limits_{0}^{2\pi} e^{cos\theta} \cdot e^{i(sin\theta - n\theta)}d\theta \\ = \int\limits_{0}^{2\pi} e^{cos\theta + isin\theta} \cdot e^{-n\theta}d\theta \\ I = \int\limits_{0}^{2\pi} e^{e^{i\theta}} \cdot e^{-in\theta}d\theta......(1)$$

Put $z = e^{i\theta}$

Taking log on both sides, we get

$logz = i\theta loge \\ \frac{1}{z} dz = id\theta \\ \therefore d\theta = \frac{dz}{iz} \\ 0 \leq \theta \leq 2\pi \hspace{1cm} \therefore C: |z| = 1$

$\therefore$ equation (1) becomes

$I = \int e^z \cdot \frac{z^{-n}dz}{iz} \\ \therefore I = \frac{1}{i} \int \bigg(\frac{e^z}{z^{n + 1}}\bigg)dz.......(2)$

$f(z) = \bigg(\frac{e^z}{z^{n + 1}}\bigg)$ has pole of order (n + 1) at z = 0

Hence $I = 2\pi i$ (Sum of the residues)...............(3)

$\therefore r_{1} = Res(z = 0) = \frac{1}{n!} \lim_{z \rightarrow 0} \frac{d^n}{dz^n} z^{n + 1} \cdot \frac{e^z}{z^{n + 1}} \\ = \frac{1}{n!} \lim_{z \rightarrow 0} (1)^n e^z = \frac{1}{n!}$

Hence $I = 2\pi i \cdot \frac{1}{i} \cdot \frac{1}{n!} = \frac{2\pi}{n!}.....$from (2) and (3)

i.e. $I = \int\limits_{0}^{2\pi} e^{cos\theta} \cdot e^{i(sin\theta - n\theta)} d\theta = \int\limits_{0}^{2\pi} e^{cos\theta} \cdot [cos(sin\theta - n\theta) + isin(sin\theta - n\theta)]d\theta = \frac{2\pi}{n!}$

Equating real and imaginary parts we get

$$\int\limits_{0}^{2\pi} e^{cos\theta} cos(sin\theta - n\theta)d\theta = \frac{2\pi}{n!}....required \hspace{0.25cm} result \\ \int\limits_{0}^{2\pi} e^{cos\theta} sin(sin\theta - n\theta)d\theta = 0$$

Hence $$\int\limits_{0}^{2\pi} e^{cos\theta} cos(sin\theta - n\theta)d\theta = \frac{2\pi}{n!}$$