Answer
$$\frac{1}{2}(x-\sin x)+c$$
Work Step by Step
Given $$\int \sin^2(x/2)dx$$
Since $$ \cos 2x=1-2\sin^2 x=1+2\cos^2x$$
Then
\begin{align*}
\int \sin^2(x/2)dx&= \frac{1}{2}\int(1- \cos(x))dx\\
&=\frac{1}{2}(x-\sin x)+c
\end{align*}
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 280: 65
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