Answer
$\dfrac{1}{8}(8-4\pi+\pi^2)$
Work Step by Step
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{x}x\sin y\,dy\,dx=-\int_{0}^{\frac{\pi}{2}}x\biggl[\cos y\biggr]_{0}^{x}\, dx=-\int_{0}^{\frac{\pi}{2}}x\cos x-x\, dx=\dfrac{1}{8}(8-4\pi+\pi^2)$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 15 - Multiple Integrals - 15.2 Double Integrals over General Regions - 15.2 Exercises - Page 1048: 4
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