Answer
$$\pi$$
Work Step by Step
\begin{align*}
\int_{0}^{2\pi}\sin^2 xdx&= \frac{1}{2} \int_{0}^{2\pi} \left( 1-\cos 2x\right) dx \\
&= \frac{1}{2}\left(x -\frac{1}{2}\sin 2x\right) \bigg|_{0}^{2\pi}\\
&=\pi
\end{align*}
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 49
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