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Would the solution to the following integral: $$\int_{|z|=1} e^{2z} \frac1z \,dz$$

be 2$\pi$i when using Cauchy's Integral Formula?

Also, how would I use the value from this integral to evaluate the following: $$\int_0^{2𝜋} e^{2\cosθ}\cos(2\sinθ) \,dθ$$

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Yes, the solution would be $2\pi i$. Letting $z=e^{i\theta}$, we have $\text{d}z=ie^{i\theta}\text{d}\theta$, and your integral becomes

$$i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta-\int_{0}^{2\pi}e^{2\cos\theta}\sin\left(2\sin\theta\right)\text{d}\theta.$$

Notice that the right integral equals $0$ because it is odd with a period of $2\pi$. Thus,

$$\int_{\left\lvert z\right\rvert=1}\frac{e^{2z}}{z}\text{d}z=i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi i,$$

giving $$\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi.$$

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