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: 48

Answer

$$\frac{1}{4}$$

Work Step by Step

Given $$ \int_{0}^{\pi / 4} x \sin 2 x d x$$ Let \begin{align*} u&=x\ \ \ \ \ \ \ \ dv=\sin 2x\\ du&=dx\ \ \ \ \ \ \ \ v= \frac{-1}{2}\cos 2x \end{align*} Then \begin{align*} \int_{0}^{\pi / 4} x \sin 2 x d x&= \frac{-x}{2}\cos 2x \bigg|_{0}^{\pi/4}+\frac{1}{2}\int_{0}^{\pi / 4} \cos2 x d x\\ &= \frac{-x}{2}\cos 2x \bigg|_{0}^{\pi/4}+ \frac{1}{4}\sin 2x\bigg|_{0}^{\pi/4}\\ &= \frac{1}{4} \end{align*}
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