# Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 50

Find the definite integral $\displaystyle \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt$

Let $\displaystyle u = \frac{2 \pi t}{T} - \alpha$, then $du = \frac{2 \pi}{T} dt$, so $\displaystyle dt = \frac{T}{2 \pi} du$. When $t = 0, u = - \alpha$ and when $\displaystyle t = \frac{T}{2}, u = \pi - \alpha$. Thus,

\begin{aligned} \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \int^{\frac{T}{2}}_0 \sin u \frac{T}{2 \pi} du \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{2 \pi} \int^{\frac{T}{2}}_0 \sin u du \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{2 \pi} \left. - \cos u \right|^{\frac{T}{2}}_0 \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{2 \pi} \left[ - \cos (\pi - \alpha) + \cos (-\alpha) \right] \qquad \text{ Apply sum and difference formula \cos (a - b) = \cos a \cos b - \sin a \sin b} \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{2 \pi} [\sin (\pi) \sin(- \alpha) - \cos(\pi) \cos (- \alpha) + \cos (- \alpha)] \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& [(0) \sin(- \alpha) - (-1) \cos (- \alpha) + \cos ( - \alpha)] \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{2 \pi} [\cos (- \alpha) + \cos (- \alpha)] \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T}{\cancel {2} \pi} \cancel{2} \cos (- \alpha) \\ \\ \int^{\frac{T}{2}}_0 \sin \left( \frac{2 \pi t}{T} - \alpha \right) dt =& \frac{T \cos (- \alpha)}{\pi} \end{aligned}

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