Why is $\int_{-1}^1 (1-|t|) \cdot e^{-i \omega t} dt$ equal to $2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt$ I have a task where I should calculate the fourier transform of
$$
\Delta(t) = \begin{cases}
(1-|t|)& |t| \le 1 \\
0 & |t| > 1
\end{cases}
$$
The solution says, that
$$
 \int_{-1}^1 (1-|t|) \cdot e^{-i\omega t} dt = 2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt
$$
I understand that the factor must be $2$ and that we only need to integrate from 0 to 1 because $\Delta(t)$ is an even function. But i don't get the last transformation.
$$
 \int_{-1}^1 (1-|t|) \cdot e^{-i\omega t} dt = 2 \cdot \int_0^1 (1-t) \cdot e^{-i\omega t}\,dt = \mathop{???} = 2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt
$$
Does anybody know how the transformation from $e^{-i\omega t}$ to $\cos{\omega t}$ works?
I would say that this are different functions, because normally i would say $e^{-i\omega t} = \cos{\omega t} - i \sin{\omega t}$
 A: They have sort of buried the idea; $\cos(\omega t)$ is an even function of $t$ and $\sin(\omega t)$ is an odd function of $t$. If you use Euler's formula first before folding the integral in half it makes more sense.
To be more specific:
\begin{align*}
\int_{-1}^{1} (1-|t|)e^{-i\omega t}dt &= \int_{-1}^{1} (1-|t|)(\cos(\omega t)-i\sin(\omega t))dt \\ &= \int_{-1}^{1} (1-|t|)\cos(\omega t)dt - i\int_{-1}^{1} (1-|t|)\sin(\omega t)dt \\ &= \int_{-1}^{1} (1-|t|)\cos(\omega t)dt +0
\end{align*}
A: You wrote

$$
\int_{-1}^1 (1-|t|) e^{-i\omega t} dt = 2\int_{0}^1 (1-t) e^{-i\omega t}dt
$$

This is false, as $e^{-i\omega t}$ is not an even function. But all functions can be divided into an even part and an odd part. For $e^{-i\omega t}$, this is given by $e^{-i\omega t} = \cos(\omega t) -i\sin(\omega t)$. So we have
$$
\int_{-1}^1 (1-|t|) e^{-i\omega t} dt = \int_{-1}^1 (1-|t|) \cos(\omega t) dt - i\int_{-1}^1 (1-|t|) \sin(\omega t) dt \\ = 2\int_{-1}^1 (1-t) \cos(\omega t) dt,
$$
where the sine integral is zero because its integrand is odd.
