I will try to provide you with some hints how to implement your unitary operation.
Let's firstly rewrite it as $U_d = \mathrm{exp}(-i\Theta(t)\sigma_x)$ or $U_d = \mathrm{exp}(-i\Theta(t)X)$ (i.e. I replaced the notation $\sigma_x$ for Pauli $X$ gate by symbol $X$).
This means that $U_d$ is in fact $x$ rotation given by formula
$$
Rx(\theta)= \mathrm{exp}\Big(-i\frac{\theta}{2}X\Big)=
\begin{pmatrix}
\cos(\theta/2) & -i\sin(\theta/2)\\
-i\sin(\theta/2) & \cos(\theta/2)
\end{pmatrix}.
$$
Setting $\theta/2 = \pi/2$ (or $\theta = \pi$) we get
$$
Rx(\pi) =
\begin{pmatrix}
\cos(\pi/2) & -i\sin(\pi/2)\\
-i\sin(\pi/2) & \cos(\pi/2)
\end{pmatrix}
=
\begin{pmatrix}
0 & -i\\
-i & 0
\end{pmatrix}
= -iX,
$$
or in other words $x$ rotation with the rotation angle equal to $\pi$ is equivalent to Pauli $X$ gate up to global phase $-i$.
So, if you are able to compute your definite integral $\Theta(t)$, then the operation $U_d$ is simply $Rx(\theta)$ gate with the angle $\theta = 2\Theta(t)$. If $2\Theta(t) = \pi$ then you will get $X$ gate up to global phase $-i$.