Answer
$\int sinudu$
$u=t^2$
Work Step by Step
Try it:
$u=t^2$
$du=2tdt$
$\int sinudu=-cosu+C$
$\frac1 2 \int sin(t^2)2tdt$
$\frac1 2 \int sinudu$
$\frac 1 2 (-cosu+C)$
$\frac 1 2 (-cos(t^2)+C)$
$\frac 1 2 -cos(t^2)+C$
$\int sinudu$
$u=t^2$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 11
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