## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises - Page 391: 55

#### Answer

$= \frac{\theta }{2} - \frac{1}{4}\sin \,\left( {2\theta + \frac{\pi }{3}} \right) + C$

#### Work Step by Step

$\begin{gathered} \int_{}^{} {{{\sin }^2}\,\left( {\theta + \frac{\pi }{6}} \right)} \,d\theta \hfill \\ \hfill \\ use\,\,the\,identity\,\,{\sin ^2}x = \frac{{1 - \cos 2x}}{2} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{{1 - \cos \,\left( {2\theta + \frac{\pi }{3}} \right)}}{2}} \,d\theta \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \int_{}^{} {\frac{1}{2}\,d\theta } - \frac{1}{4}\int_{}^{} {\cos \,\left( {2\theta + \frac{\pi }{3}} \right)\,\left( 2 \right)\,d\theta } \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{\theta }{2} - \frac{1}{4}\sin \,\left( {2\theta + \frac{\pi }{3}} \right) + C \hfill \\ \end{gathered}$

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