Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 10

Answer

$$\frac{\pi ^{\frac{3}{2}}}{12}$$

Work Step by Step

Given $$ \int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}dx$$ Let $u=\tan^{-1}x \ \ \Rightarrow \ \ du =\dfrac{1}{1+x^2}dx $, and at $x=0 \to u=0$, $x=1 \to u=\pi/4$, then \begin{align*} \int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}dx&= \int_{0}^{\pi/4} \sqrt{u}du\\ &=\frac{2}{3}u^{3/2}\bigg|_{0}^{\pi/4}\\ &=\frac{2}{3}\left(\frac{\pi}{4}\right)^{3/2}\\ &=\frac{\pi ^{\frac{3}{2}}}{12} \end{align*}
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