Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 524: 10


$$\frac{\pi }{16}$$

Work Step by Step

Given $$ \int_{0}^{\pi}\sin^2 t\cos^4tdt$$ since $$\cos^2x=\frac{1}{2}(1+\cos 2x),\ \ \ \sin^2x=\frac{1}{2}(1-\cos 2x)$$ then \begin{align*} \int_{0}^{\pi}\sin^2 t\cos^4( t)dt&=\frac{1}{8}\int_{0}^{\pi}[ (1+\cos 2t)^2 (1-\cos 2t)]dt\\ &=\frac{1}{8}\int_{0}^{\pi}[ (1+\cos 2t) (1-\cos^2 2t)]dt\\ &=\frac{1}{8}\int_{0}^{\pi}[ 1-\cos^2 2t + \cos 2t-\cos^3 2t)]dt\\ &=\frac{1}{8}\int_{0}^{\pi}[ \frac{1}{2}-\frac{1}{2}\cos4t + \cos 2t-(1-\sin^2(2t))\cos 2t)]dt\\ &=\frac{1}{8}\int_{0}^{\pi}[ \frac{1}{2}-\frac{1}{2}\cos4t + \sin^2(2t) \cos 2t)]dt\\ &=\frac{1}{8}\left( \frac{1}{2}t-\frac{1}{8}\sin4t + \frac{1}{6} \sin^3(2t) \right)\bigg|_{0}^{\pi}\\ &=\frac{\pi }{16} \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.