## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises: 43

#### Answer

$$\int^{1/\sqrt3}_{0}\frac{t^2-1}{t^4-1}dt=\frac{\pi}{6}$$

#### Work Step by Step

$$A=\int^{1/\sqrt3}_{0}\frac{t^2-1}{t^4-1}dt$$ $$A=\int^{1/\sqrt3}_{0}\frac{t^2-1}{(t^2-1)(t^2+1)}dt$$ $$A=\int^{1/\sqrt3}_{0}\frac{1}{t^2+1}dt$$ According to Table 1, we have $$\int\frac{1}{x^2+1}dx=\tan^{-1}x+C$$ Therefore, $$A=(\tan^{-1}x)\Bigg]^{1/\sqrt 3}_{0}$$ $$A=\tan^{-1}(\frac{1}{\sqrt 3})-\tan^{-1}0$$ We know that $\tan^{-1}(\frac{1}{\sqrt 3})=\frac{\pi}{6}$. For $\tan^{-1}0$, we see that as $\tan x=0$, $\sin x$ must also equal $0$, since $\tan x=\frac{\sin x}{\cos x}$. And we have $\sin0 =0$, so $\tan^{-1}0=0$. $$A=\frac{\pi}{6}-0$$ $$A=\frac{\pi}{6}$$

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.