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: 32

Answer

$$\int\frac{\sqrt{2-x}}{\sqrt{x}}dx=\sqrt{x(2-x)}+2\sin^{-1}\sqrt{\frac{x}{2}}+C$$

Work Step by Step

$$A=\int\frac{\sqrt{2-x}}{\sqrt{x}}dx$$ We set $u=\sqrt x$, which means $$du=\frac{1}{2\sqrt x}dx$$ $$\frac{1}{\sqrt{x}}dx=2du$$ Also, we can rewrite $x=u^2$ Therefore, $$A=2\int\sqrt{2-u^2}du$$ Here apply Formula 45, which states that $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$$ for $a=\sqrt2$. $$A=2\Big(\frac{u}{2}\sqrt{2-u^2}+\frac{2}{2}\sin^{-1}\frac{u}{\sqrt2}\Big)+C$$ $$A=u\sqrt{2-u^2}+2\sin^{-1}\frac{u}{\sqrt2}+C$$ $$A=\sqrt x\sqrt{2-x}+2\sin^{-1}\frac{\sqrt x}{\sqrt2}+C$$ $$A=\sqrt{x(2-x)}+2\sin^{-1}\sqrt{\frac{x}{2}}+C$$
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