Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 45


$$\frac{1}{2}\left(\sin t^2 -\frac{1}{3}\sin^3 t^2\right)+c$$

Work Step by Step

Given $$\int t\cos^3t^2dt $$ Let $$u=t^2\ \ \ \Rightarrow \ \ du=2tdt$$ Then \begin{align*} \int t\cos^3t^2dt&=\frac{1}{2}\int \cos^3udu\\ &=\frac{1}{2}\int \cos^2u\cos u du\\ &=\frac{1}{2}\int (1-\sin^2u)\cos u du\\ &=\frac{1}{2}\left(\sin u -\frac{1}{3}\sin^3 u\right)+c\\ &=\frac{1}{2}\left(\sin t^2 -\frac{1}{3}\sin^3 t^2\right)+c \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.