University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 31



Work Step by Step

$$A=\int^{\pi/2}_0\theta\sqrt{1-\cos2\theta}d\theta$$ Use the identity: $$\frac{1-\cos2\theta}{2}=\sin^2\theta$$ $$1-\cos2\theta=2\sin^2\theta$$ Therefore, $$A=\int^{\pi/2}_0\theta\sqrt{2\sin^2\theta}d\theta$$ $$A=\sqrt2\int^{\pi/2}_0\theta\sin\theta d\theta$$ (since on $[0,\pi/2]$, $\sin\theta\ge0$) Now we apply Integration by Parts. Set $u=\theta$ and $dv=\sin\theta d\theta$ So we would have $du=d\theta$ and $v=-\cos\theta$ According to Integration by Parts rule, we have $$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0-\int^{\pi/2}_0-\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0+\int^{\pi/2}_0\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\frac{\pi}{2}\cos\frac{\pi}{2}-0+(\sin\theta]^{\pi/2}_0)\Big)$$ $$A=\sqrt2\Big(0+\sin\frac{\pi}{2}-\sin0\Big)$$ $$A=\sqrt2(1)=\sqrt2$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.