## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 15

#### Answer

$$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C$$

#### Work Step by Step

\eqalign{ & \int {{{\sin }^{ - 1}}x} dx \cr & {\text{substitute }}u = {\sin ^{ - 1}}x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & dv = dx,{\text{ }}v = x \cr & {\text{using integration by parts}} \cr & \int {udv} = uv - \int {vdu} \cr & {\text{, we have}} \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx} \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x\left( {\frac{1}{{{{\left( {1 - {x^2}} \right)}^{1/2}}}}} \right)dx} \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x{{\left( {1 - {x^2}} \right)}^{ - 1/2}}dx} \cr & {\text{find antiderivative}}{\text{, }}t = 1 - {x^2},{\text{ }}dt = - 2xdx \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \frac{1}{2}\int {{t^{ - 1/2}}dt} \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \frac{1}{2}\frac{{{t^{1/2}}}}{{1/2}} + C \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + {t^{1/2}} + C \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + {\left( {1 - {x^2}} \right)^{1/2}} + C \cr & \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C \cr}

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