Answer
$$\int_0^{T/2}\sin\left(\frac{2\pi t}{T}-\alpha\right)dt=\frac{T}{\pi}\cos\alpha$$
Work Step by Step
To evaluate the integral $$\int_0^{T/2}\sin\left(\frac{2\pi t}{T}-\alpha\right)dt$$
we will use substitution $\frac{2\pi t}{T}-\alpha=z$ which gives us $\frac{2\pi}{T}dt=dz\Rightarrow dt=\frac{T}{2\pi}dz$. New integration bounds would be: for $t=0$ we have $z=-\alpha$ and for $t=T/2$ we have $z=\pi-\alpha$. Putting this into the integral we get:
$$\int_0^{T/2}\sin\left(\frac{2\pi t}{T}-\alpha\right)dt=
\int_{-\alpha}^{\pi-\alpha}\sin z\frac{T}{2\pi}dz=
\frac{T}{2\pi}\int_{-\alpha}^{\pi-\alpha}\sin zdz=
\frac{T}{2\pi}\left.(-\cos z\right|_{-\alpha}^{\pi-\alpha})=
-\frac{T}{2\pi}(\cos(\pi-\alpha)-\cos(-\alpha))=
-\frac{T}{2\pi}(-2\cos\alpha)=\frac{T}{\pi}\cos\alpha$$
