Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 82

Answer

\[ = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} \hfill \\ \hfill \\ integration\,\,by\,\,parts\,\,formula \hfill \\ \hfill \\ \int_{}^{} {udv = uv - \int_{}^{} {vdu} } \hfill \\ \hfill \\ set \hfill \\ u = {\sin ^{ - 1}}x\,\,\,\,then\,\,\,du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \hfill \\ \hfill \\ dv = {x^n}dx\,\,\,\,\,\,then\,\,\,\,\,\,\,v = \frac{{{x^{n + 1}}}}{{n + 1}} \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \int_{}^{} {\frac{{{x^{n + 1}}}}{{n + 1}}} \,\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx \hfill \\ \hfill \\ = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \end{gathered} \]
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