Calculus 8th Edition

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

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 42


$$\int_0^{\pi/2}\cos x\sin(\sin x)dx=1-\cos1$$

Work Step by Step

To evaluate the integral $$\int_0^{\pi/2}\cos x\sin(\sin x)dx$$ we will use substitution $\sin x=t$ which gives us $\cos xdx=dt$ and the integration bounds would be: for $x=0$ we have $t=0$ and for $x=\pi/2$ we have $t=1$ so we have: $$\int_0^{\pi/2}\cos x\sin(\sin x)dx=\int_0^1\sin tdt=\left.-\cos t\right|_0^1=-(\cos1-\cos0)=-(\cos1-1)=1-\cos1$$
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