University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 33


$$\int\cot t\sqrt{1-\sin^2t}dt=\sqrt{1-\sin^2t}-\ln\Big|\frac{1+\sqrt{1-\sin^2t}}{\sin t}\Big|+C$$

Work Step by Step

$$A=\int\cot t\sqrt{1-\sin^2t}dt$$ ($0\lt t\lt\pi/2$) $$A=\int\frac{\cos t}{\sin t}\sqrt{1-\sin^2t}dt$$ We set $u=\sin t$, which means $$du=\cos tdt$$ Therefore, $$A=\int\frac{\sqrt{1-u^2}}{u}du$$ Here apply Formula 47, which states that $$\int\frac{\sqrt{a^2-x^2}}{x}dx=\sqrt{a^2-x^2}-a\ln\Big|\frac{a+\sqrt{a^2-x^2}}{x}\Big|+C$$ for $a=1$. $$A=\sqrt{1-u^2}-\ln\Big|\frac{1+\sqrt{1-u^2}}{u}\Big|+C$$ $$A=\sqrt{1-\sin^2t}-\ln\Big|\frac{1+\sqrt{1-\sin^2t}}{\sin t}\Big|+C$$
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