Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 25

Answer

$$\int arccos\,x\,dx=x\,arccos\,x-\sqrt{1-x^{2}}+C$$

Work Step by Step

${(x\,arccos\,x)}'=arccos\,x+\frac{x}{\sqrt{1-x^{2}}}$ $$\int arccos\,x\,dx=x\,arccos\,x+\int \frac{x}{\sqrt{1-x^{2}}}dx$$ $For\ \int \frac{x}{\sqrt{1-x^{2}}}dx,\ let\ x=sin\,t,dx=cos\,t\,dt.$ $$\int \frac{x}{\sqrt{1-x^{2}}}dx=\int \frac{sin\,t}{cos\,t}cos\,t\,dt$$ $$=-cos\,t+C=-\sqrt{1-x^{2}}+C$$ $$\int arccos\,x\,dx=x\,arccos\,x-\sqrt{1-x^{2}}+C$$
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