Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.4 Exercises - Page 539: 32

Answer

$$\arcsin \sqrt x + \sqrt {x\left( {1 - x} \right)} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sqrt {1 - x} }}{{\sqrt x }}} dx \cr & {\text{Rewrite}} \cr & \int {\frac{{\sqrt {1 - x} }}{{\sqrt x }}} dx = \int {\frac{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}{{\sqrt x }}} dx \cr & {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}dx = 2\sqrt x du \cr & \int {\frac{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}{{\sqrt x }}} dx = \int {\frac{{\sqrt {1 - {u^2}} }}{{\sqrt x }}} \left( {2\sqrt x } \right)du \cr & = 2\int {\sqrt {1 - {u^2}} } du \cr & {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr & \int {\sqrt {{a^2} - {u^2}} du = \frac{1}{2}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C} \cr & 2\int {\sqrt {1 - {u^2}} } du = {\left( 1 \right)^2}\arcsin \frac{{\sqrt x }}{1} + \sqrt x \sqrt {1 - {{\left( {\sqrt x } \right)}^2}} + C \cr & {\text{Simplifying}} \cr & \arcsin \sqrt x + \sqrt x \sqrt {1 - x} + C \cr & \arcsin \sqrt x + \sqrt {x\left( {1 - x} \right)} + C \cr} $$
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