## Calculus 8th Edition

Published by Cengage

# Chapter 7 - Techniques of Integration - 7.3 Trigonometric Substitution - 7.3 Exercises - Page 531: 22

#### Answer

$$\displaystyle{\int_{0}^1\sqrt{x^2+1}dx=\frac{\sqrt{2}+\ln\left(1+\sqrt{2}\right)}{2}}\\$$

#### Work Step by Step

$\displaystyle{I=\int_{0}^1\sqrt{x^2+1}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=\tan\theta & x^2=\tan^2\theta \\ & \\ \frac{dx}{d\theta}=\sec^2\theta & dx=\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_{0}^{\frac{\pi}{4}}\sqrt{\tan^2\theta+1}\sec^2\theta\ d\theta}\\ \displaystyle{I=\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta}\\$ $\displaystyle \left[\begin{array}{ll} u=\sec\theta & dv=\sec^2\theta \\ & \\ du=\sec\theta\tan\theta & v=\tan\theta \end{array}\right]$ Integration by parts $\displaystyle{\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\left[\sec\theta\tan\theta\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\sec\theta\tan^2\theta\ d\theta}\\ \displaystyle{\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\sqrt{2}-\int_{0}^{\frac{\pi}{4}}\sec\theta\left(\sec^2\theta-1\right)\ d\theta}\\ \displaystyle{\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\sqrt{2}-\int_{0}^{\frac{\pi}{4}}\sec^3\theta-\sec\theta\ d\theta}\\ \displaystyle{\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\sqrt{2}-\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta+\int_{0}^{\frac{\pi}{4}}\sec\theta\ d\theta}\\ \displaystyle{2\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\sqrt{2}+\int_{0}^{\frac{\pi}{4}}\sec\theta\ d\theta}\\ \displaystyle{\int_{0}^{\frac{\pi}{4}}\sec^3\theta\ d\theta=\frac{1}{2}\left(\sqrt{2}+\left[\ln\left|\sec \theta+\tan \theta\right|\right]_{0}^{\frac{\pi}{4}}\right)}\\ \displaystyle{I=\frac{1}{2}\left(\sqrt{2}+\left(\ln\left(1+\sqrt{2}\right)-\ln1\right)\right)}\\ \displaystyle{I=\frac{\sqrt{2}+\ln\left(1+\sqrt{2}\right)}{2}}\\$

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