Cauchy's Integral Question Complex Number

Question

I have a question and I'm kind of stuck, I was wondering if you were able to help me move forward.

The question is, Use Cauchy's integral formula to evaluate,

$$ \int_{|z| = 1}\frac{e^{2z}}{z^2}dz $$

where $|z| = 1$ is oriented counter-clockwise. Using this result, evaluate the real integral

$$ \int_0^{2\pi}e^{2\cos(t)}\cos(2\sin(t) - t)dt $$

I have gotten to this step, and I'm stuck. I have uploaded an image. Please do let me know if its unclear, ill rewrite it and upload another picture because Im not familiar with LaTex. Here's my work:

$$ |z| = 1 \text{ as } z(t) = e^{it}, t\in [0, 2\pi] $$

\begin{align} \int_{|z|} =&\ \int_0^{2\pi}\frac{e^{2e^{it}}}{\left(e^{it}\right)^2}ie^{it}dt \\ =&\ i\int_0^{2\pi}e^{-it}e^{2\cos(t) + 2i\sin(t)}dt \\ =&\ i\int_0^{2\pi} e^{2\cos(t) + i(2\sin(t) - t)}dt \end{align}

Answer 1

$$f(z)=e^{2z}$$

Using:

$f'(z_0)=$

$$f'(0)=\frac{1}{2 \pi i}\int _\gamma \frac{e^{2z}}{z^2}dz$$

$$f'(0)=2$$

The answer becomes: $$4 \pi i$$

Answer 2

HINT: Remember that $$e^{i(2 \sin t-t)} = \cos(2\sin t-t)+i\sin(2\sin t-t)$$

Follow the substitution through and then consider real and imaginary parts.

Answer 3

Actually, you are using a method when $cos \theta$ or $sin \theta$ should be in the integration.The simple method is either use cauchy integral formula or cauchy residue theorem.

$f^{(n-1)}(z_0) =\frac{1}{2 \pi i} \int_C \frac{f(z)dz}{(z-z_0)^n}$

Comparing this with your integration, $f(z) = e^{2z}, z_0 = 0, n=2$ and $C$ is $|z| = 1$.

check whether $z_0 = 0$ lies within $C$. (yes) so apply the above formula. Find $f'(z) = 2 e^{2z}$ and substitute $z = 0$. we get $2e^0$. the answer is $2.2 \pi i$ = $4 \pi i. $