Derivation of $\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)\cos(t\xi)dt$ 
Problem: Consider the Fourier transform: $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{-it\xi}dt\quad\text{and}\quad f\in{L_1}_{(\mathbb{R})}$$
Now it was defined to me on ${L_1}_{(\mathbb{R})}$ $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)\cos(t\xi)dt$$

Given the Euler identity $e^{it\xi}=\cos(t\xi)+i\sin(t\xi)$, I do not know how the $i\sin(t\xi)$ vanished.
Question:
How did it happen?
Thanks in advance!
Edit: Urgent
I only noticed it now, the transform defined is the following:
$$\hat{f}(\xi)=\frac{\sqrt{2}}{\sqrt{\pi}}\int_\limits{0}^{\infty}f(t)\cos(t\xi)dt$$
Apologies!
 A: That identity can only work for even functions. Recall that if a function $a(t)$ is odd, then $\int_{-t}^t a(\tau) d\tau = 0$.
Thus, for even $f$, $$\int_{-\infty}^\infty f(t) \sin(\xi t) dt = 0$$ since $f(t) \sin(\xi t)$ is odd for all $\xi$. (Product of an even function with an odd function is odd.)

Now, suppose that $\hat f(\xi) = \int_{-\infty}^\infty f(t) \cos(\xi t) dt.$ We have that $\hat f$ is even, since $\cos$ is even.
If we want to recover $f$, then we can apply the inverse Fourier transform: $$f(t) = \int_{-\infty}^\infty \hat f(\xi) e^{i\xi t} d\xi = \int_{-\infty}^\infty \hat f(\xi) \cos(\xi t) d\xi + i \int_{-\infty}^\infty \hat f(\xi) \sin(\xi t) d\xi.$$ Just as before $\hat f$ even means that $\int_{-\infty}^\infty \hat f(\xi) \sin(\xi t) d\xi  = 0$. Hence, $f(t) = \int_{-\infty}^\infty \hat f(\xi) \cos(\xi t) d\xi$.
Finally, we have that $f$ is even, since $\cos$ is even.
Thus, $\hat f(\xi) = \int_{-\infty}^\infty f(t) \cos(\xi t) dt$ iff $f$ is even.

As for your edit, I did neglect the constants above. This is often done for expediency, and I can see how it can be confusing.
Taking $\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{-\xi t} dt$, we already established that $\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) \cos(\xi t) dt$.
Now for even functions, say $b(t)$, we have $\int_{-a}^a b(t) dt = 2 \int_0^a b(t) dt$. Thus, $$\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) \cos(\xi t) dt = \frac{2}{\sqrt{2\pi}} \int_{0}^\infty f(t) \cos(\xi t) dt = \frac{\sqrt2}{\sqrt{\pi}} \int_{0}^\infty f(t) \cos(\xi t) dt$$
