Answer
$-1$
Work Step by Step
$\int_{\pi/2}^{\pi} (\frac{sin(2x)}{2\cdot sinx}) dx=\int_{\pi/2}^{\pi} (\frac{2sinxcosx}{2\cdot sinx}) dx=\int_{\pi/2}^{\pi} (cosx) dx=[sinx]_{\pi/2}^{\pi}=[sin\pi]-[sin\frac{\pi}{2}]=0-1=-1$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 25
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