Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 50


$$\int_0^{T/2}\sin\left(\frac{2\pi t}{T}-\alpha\right)dt=\frac{T}{\pi}\cos\alpha$$

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: $$\int_0^{T/2}\sin\left(\frac{2\pi t}{T}-\alpha\right)dt= \int_{-\alpha}^{\pi-\alpha}\sin z\frac{T}{2\pi}dz= \frac{T}{2\pi}\int_{-\alpha}^{\pi-\alpha}\sin zdz= \frac{T}{2\pi}\left.(-\cos z\right|_{-\alpha}^{\pi-\alpha})= -\frac{T}{2\pi}(\cos(\pi-\alpha)-\cos(-\alpha))= -\frac{T}{2\pi}(-2\cos\alpha)=\frac{T}{\pi}\cos\alpha$$
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