So, in order to calculate the Fourier Transform of: $$x(t) = \sin(\omega_0t)\times\Pi(\frac{t-\frac{T}{2}}{T})$$
I've begun by calculating the Fourier Transform of each of the factors.
According to this Fourier Transform pair table:
$$A(\omega) = \mathcal{F}\{\sin(\omega_0t)\}=\frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)] = -\pi j[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$
And since the second factor can be written as: $$\Pi(\frac{t-\frac{T}{2}}{T})= \begin{cases} 1, & 0<t<T \\ 0, & otherwise \end{cases}$$
Its Fourier Transform can be calculated by:
$$B(\omega)= \mathcal{F}\{\Pi(\frac{t-\frac{T}{2}}{T})\} = \int_{-\infty}^\infty \Pi(\frac{t-\frac{T}{2}}{T})\times e^{-j\omega t}\, dt = \int_0^T 1\times e^{-j\omega t}\, dt = \dots = \frac{1-e^{-j\omega T}}{j\omega}$$
Therefore, since multiplication in the time domain is equal to convolution in the frequency domain:
$$\mathcal{F}\{x(t)\} = \mathcal{F}\{\sin(\omega_0t)\times\Pi(\frac{t-\frac{T}{2}}{T})\} = A(\omega)*B(\omega) = \{-\pi j[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\}*\{\frac{1-e^{-j\omega T}}{j\omega}\} $$
Now, after using the linearity property and the sifting property of the delta function:
$$\mathcal{F}\{x(t)\} = -\pi j \{\delta(\omega-\omega_0)*\frac{1-e^{-j\omega T}}{j\omega}-\delta(\omega+\omega_0)*\frac{1-e^{-j\omega T}}{j\omega}\} = \dots = \pi \{\frac{1-e^{j(\omega+\omega_0)T}}{\omega+\omega_0}-\frac{1-e^{j(\omega-\omega_0)T}}{\omega-\omega_0}\}$$
Is my solution correct?