Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 535: 35

Answer

$$ \int_{0}^{\pi / 2} \tan ^{2} \theta d \theta $$ We note first the given integral is improper because $ f(x)= \tan ^{2} \theta $ has the vertical asymptote $\theta =\pi / 2 $. The infinite discontinuity occurs at the right endpoint of $ [0, \pi / 2] $ The given improper integral is divergent.

Work Step by Step

$$ \int_{0}^{\pi / 2} \tan ^{2} \theta d \theta $$ Observe that the given integral is improper because $ f(x)= \tan ^{2} \theta $ has the vertical asymptote $\theta =\pi / 2 $. Since the infinite discontinuity occurs at the right endpoint of $ [0, \pi / 2] $ , we must use part (a) of Definition 3 : $$ \begin{split} \int_{0}^{\pi / 2} \tan ^{2} \theta d \theta & = \lim _{t \rightarrow \pi / 2 ^{-}} \int_{0}^{t} \tan ^{2} \theta d \theta \\ & =\lim _{t \rightarrow \pi / 2 ^{-}} \int_{0}^{t} (\ sec^{2}\theta -1) d \theta \\ & =\lim _{t \rightarrow \pi / 2 ^{-}}[\tan \theta- \theta ]_{0}^{t} \\ & =\lim _{t \rightarrow \pi / 2 ^{-}}[\tan t - t] \\ & =\infty, \end{split} $$ so the integral $$ \int_{0}^{\pi / 2} \tan ^{2} \theta d \theta $$ is divergent. Thus the given improper integral is divergent.
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