Help with calculating the integral $\int_{-\pi}^\pi \cos\left(x/2\right) \cdot e^{ix} dx$ by using Eulers formula I have to determine the following integral
$$
\int_{-\pi}^\pi \cos\left(x/2\right) \cdot e^{ix} dx
$$
by using Eulers formula
$$
\cos\left(x/2\right) = \frac{e^{ix/2}+e^{-ix/2}}{2}
$$
we have that
\begin{align*}
\int_{-\pi}^\pi \cos\left(x/2\right) e^{ix} dx & = \frac{1}{2} \int_{-\pi}^\pi \left(e^{ix/2}+e^{-ix/2}\right) e^{ix} dx \\
& = \frac{1}{2} \int_{-\pi}^\pi e^{\frac{i3x}{2}}+e^{-\frac{i3x}{2}} dx \\
& = \frac{1}{2} \left[ \frac{2}{3i} e^{\frac{i3x}{2}} - \frac{2}{3i} e^{\frac{-i3x}{2}}\right]_{-\pi}^\pi \\ & = \frac{1}{2}\left(\frac{2}{3i}\cdot \left(-i\right)-\frac{2}{3i}\cdot \left(\frac{1}{-i}\right)-\left(\frac{2}{3i}\cdot \left(\frac{1}{-i}\right)-\frac{2}{3i}\cdot \left(-i\right)\right)\right) \\
& = -4/3
\end{align*}
but the answer should be $4/3$. Why do I get a minus instead? Do you mind helping me? Thanks in advance. 
 A: You have a small error going from the first to the second line in your last formula... $e^{-ix/2} e^{ix} = e^{ix/2}$.
A: Well, as you wrote we know that (and it is really easy to prove):
$$\cos(x)=\frac{\exp\left(xi\right)+\exp\left(-xi\right)}{2}\tag1$$
Where $\exp(\cdot)$ is the Exponential function.
So, we also know that:
$$\cos\left(\text{n}x\right)=\frac{\exp\left(\text{n}xi\right)+\exp\left(-\text{n}xi\right)}{2}\tag2$$
So, for your integral we have:
$$\mathcal{I}_\text{n}\left(\beta,\text{k}\right):=\int_{-\beta}^\beta\cos\left(\text{n}x\right)\exp\left(\text{k}xi\right)\space\text{d}x\tag3$$
Using $(2)$ we can rewrite $(3)$:
$$\mathcal{I}_\text{n}\left(\beta,\text{k}\right)=\int_{-\beta}^\beta\frac{\exp\left(\text{n}xi\right)+\exp\left(-\text{n}xi\right)}{2}\cdot\exp\left(\text{k}xi\right)\space\text{d}x=$$
$$\frac{1}{2}\cdot\left\{\int_{-\beta}^\beta\exp\left(\text{n}xi\right)\exp\left(\text{k}xi\right)\space\text{d}x+\int_{-\beta}^\beta\exp\left(-\text{n}xi\right)\exp\left(\text{k}xi\right)\space\text{d}x\right\}=$$
$$\frac{1}{2}\cdot\left\{\int_{-\beta}^\beta\exp\left(\text{n}xi+\text{k}xi\right)\space\text{d}x+\int_{-\beta}^\beta\exp\left(\text{k}xi-\text{n}xi\right)\space\text{d}x\right\}=$$
$$\frac{1}{2}\cdot\left\{\int_{-\beta}^\beta\exp\left(x\left(\text{n}+\text{k}\right)i\right)\space\text{d}x+\int_{-\beta}^\beta\exp\left(x\left(\text{k}-\text{n}\right)i\right)\space\text{d}x\right\}\tag4$$
Now, we can solve another integral:
$$\int_{-\alpha}^\alpha\exp\left(\text{p}x\right)\space\text{d}x\tag5$$
Substitute $\text{u}=\text{p}x$, so we get:
$$\int_{-\alpha}^\alpha\exp\left(\text{p}x\right)\space\text{d}x=\frac{1}{\text{p}}\int_{-\alpha\text{p}}^{\alpha\text{p}}\exp\left(\text{u}\right)\space\text{du}=\frac{1}{\text{p}}\cdot\left[\exp\left(\text{u}\right)\right]_{-\alpha\text{p}}^{\alpha\text{p}}=$$
$$\frac{\exp\left(\alpha\text{p}\right)-\exp\left(-\alpha\text{p}\right)}{\text{p}}\tag6$$
So:

*

*$$\int_{-\beta}^\beta\exp\left(x\left(\text{n}+\text{k}\right)i\right)\space\text{d}x=\frac{\exp\left(\beta\left(\text{n}+\text{k}\right)i\right)-\exp\left(-\beta\left(\text{n}+\text{k}\right)i\right)}{\left(\text{n}+\text{k}\right)i}\tag7$$

*$$\int_{-\beta}^\beta\exp\left(x\left(\text{k}-\text{n}\right)i\right)\space\text{d}x=\frac{\exp\left(\beta\left(\text{k}-\text{n}\right)i\right)-\exp\left(-\beta\left(\text{k}-\text{n}\right)i\right)}{\left(\text{k}-\text{n}\right)i}\tag8$$

So, in your case we have $\beta=\pi$, $\text{n}=\frac{1}{2}$, and $\text{k}=1$, so:
$$\mathcal{I}_\frac{1}{2}\left(\pi,1\right)=\frac{4}{3}\tag9$$

