Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises - Page 301: 39

Answer

$\frac{1}{4}\sin^2{2x}+C$

Work Step by Step

$\int\sin{2x}\cos{2x}dx$ Method 1: Let $u=\sin{2x}\hspace{8mm}du=2\cos{2x}dx\hspace{8mm}dx=\frac{du}{2\cos{2x}}$ Therefore $\int\sin{2x}\cos{2x}dx=\int u\cos{2x}\frac{du}{2\cos{2x}}=\frac{1}{2}\int udu$ $=\frac{1}{2}\frac{u^2}{2}+C=\frac{1}{4}\sin^2{2x}+C$
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.