Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.8 Improper Integrals - 7.8 Exercises - Page 574: 15

Answer

Divergent

Work Step by Step

\[I=\int_{0}^{\infty}\sin^2{\alpha}\:d\alpha\] \[I=\lim_{t\rightarrow \infty}\int_{0}^{t}\sin^2{\alpha}\:d\alpha\;\;\;\ldots (1)\] Let\[I_1=\int\sin^2{\alpha}\:d\alpha\] \[\left[\sin^2 \theta=\frac{1-\cos 2\theta}{2}\right]\] \[I_1=\int\left( \frac{1-\cos 2\alpha}{2} \right)\:d\alpha\] \[I_1=\frac{\alpha}{2}-\frac{\sin 2\alpha}{4}\;\;\;\ldots(2)\] Using (2) in (1) \[I=\lim_{t\rightarrow \infty}\left[\frac{\alpha}{2}-\frac{\sin 2\alpha}{4}\right]_{0}^{t}\] \[I=\lim_{t\rightarrow \infty}\left[\frac{t}{2}-\frac{\sin 2t}{4}-0\right]\] $\;\;\;\;\;\;\;\;\; $ does not exist Since limit on R.H.S. of (1) does not exist so $I$ is divergent.
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