I have find out the Fourier transform of the $\sin(t)$ for range $0$ to $\pi$. Graph is added in the link.
https://www.mathcha.io/editor/p2g4u6yI3nHX6hWk
As $\sin(t)$ is a real and odd function, can I represent Fourier transform of $\sin(t)$ for $0$ to $\pi$ as $$ G(f) = -j\int_{-\infty}^\infty \sin(t) e^{-j\omega t}dt\\=-j\int_0^\pi \sin(t) \sin(\omega t)dt\\ \textrm{where}, ~ e^{-j\omega t} = \cos(\omega t)-j\sin(\omega t)\\ \int \sin(t)\cos(\omega t) = 0$$ as $\sin(t)$ is odd function. When I solve it I get $$ G(f) = \frac{-j}{2}\int_0^\pi [\cos(\omega-1)t - \cos(\omega +1)t] ~dt\\ = -j \frac{\sin(\omega\pi)}{1-\omega ^2} $$ but the solution answer is different $$G(f) = \frac{1 + e^{-2 f j \pi^2}}{1 + 4 f^2 j^2 \pi^2}$$
There is a mistake in this formula: $$\int \sin(t)\cos(\omega t) = 0$$ When you integrate it from $0$ to $\pi$, you should get $$\frac{\cos \omega\pi + 1}{1-\omega^2}$$ For the rest just use $\omega=2\pi f$ and $j^2=-1$.