Answer
$\int_{0}^{T/2}sin(\frac{2\pi t}{T}-\alpha)~dt = \frac{T}{2\pi}[cos~(-\alpha)-cos~(\pi-\alpha)]$
Work Step by Step
$\int_{0}^{T/2}sin(\frac{2\pi t}{T}-\alpha)~dt$
Let $u = \frac{2\pi t}{T}-\alpha$
$\frac{du}{dt} = \frac{2\pi}{T}$
$dt = \frac{T~du}{2\pi}$
When $t = 0$, then $u = -\alpha$
When $t = \frac{T}{2}$, then $u = \pi-\alpha$
$\int_{-\alpha}^{\pi-\alpha} \frac{T}{2\pi}sin~u~du$
$=\frac{T}{2\pi}(-cos~u~\vert_{-\alpha}^{\pi-\alpha})$
$=\frac{T}{2\pi}[-cos~(\pi-\alpha)-(-cos~(-\alpha))]$
$=\frac{T}{2\pi}[cos~(-\alpha)-cos~(\pi-\alpha)]$
