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: 38


$$\int_0^{\sqrt \pi}x\cos(x^2)dx=0$$

Work Step by Step

To evaluate the integral $$\int_0^{\sqrt \pi}x\cos(x^2)dx$$ we will use substitution $x^2=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$ and the integration bounds would be: for $x=0$ we have $t=0$ and for $x=\sqrt\pi$ we have $t=\pi$. Putting this into the integrate we get: $$\int_0^{\sqrt \pi}x\cos(x^2)dx=\int_0^\pi\cos t\frac{dt}{2}=\frac{1}{2}\left.\sin t\right|_0^\pi=\frac{1}{2}(\sin \pi -\sin0)=\frac{1}{2}(0-0)=0$$
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