Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 17

Answer

$$\frac{\pi^2}{4}$$

Work Step by Step

Given $$\int_{0}^{\pi} t\cos^2tdt$$ Since \begin{align*} \int_{0}^{\pi} t\cos^2tdt&=\frac{1}{2}\int_{0}^{\pi} t[1+\cos2t]dt\\ &=\frac{1}{2}\int_{0}^{\pi} tdt+\frac{1}{2}\int_{0}^{\pi}t\cos2t dt\\ &=\frac{1}{4} t^2 \bigg|_{0}^{\pi}+\frac{1}{2}\int_{0}^{\pi}t\cos2t dt\\ \end{align*} To evaluate $\displaystyle\int_{0}^{\pi}t\cos2t dt$ Let \begin{align*} u&=t \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos2t dt\\ du&=dt \ \ \ \ \ \ \ \ \ \ \ \ v=\frac{1}{2}\sin2t \end{align*} Then \begin{align*} \int_{0}^{\pi}t\cos2t dt&=\frac{1}{2}t\sin2t \bigg|_{0}^{\pi} -\frac{1}{2}\int_{0}^{\pi}\sin2t dt\\ &=\frac{1}{4} \cos 2t\bigg|_{0}^{\pi}\\ &=0 \end{align*} Hence $$\int_{0}^{\pi} t\cos^2tdt=\frac{\pi^2}{4} $$
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