## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 16

#### Answer

$$x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt {1 - 4{x^2}} + C$$

#### Work Step by Step

\eqalign{ & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx \cr & {\text{substitute }}u = {\cos ^{ - 1}}\left( {2x} \right),{\text{ }}du = - \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}dx \cr & du = - \frac{2}{{\sqrt {1 - 4{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 {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \int {x\left( { - \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}} \right)dx} \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) + 2\int {\frac{x}{{\sqrt {1 - 4{x^2}} }}dx} \cr & {\text{find antiderivative}}{\text{, }}t = 1 - 2{x^2},{\text{ }}dt = - 8xdx \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) + 2\int {\frac{{\left( { - 1/8} \right)dt}}{{\sqrt t }}} \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\int {{t^{ - 1/2}}} dt \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\left( {2{t^{1/2}}} \right) + C \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt t + C \cr & {\text{replace }}t = 1 - 4{x^2} \cr & \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt {1 - 4{x^2}} + C \cr}

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