Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises: 7

Answer

$\left(x^2+2x\right)\left(sinx\right)+\left(2x+2\right)cosx-2sinx+C$

Work Step by Step

Let $u=x^2+2x$ , $dv=cosx$ $du=2x+2$ , $v=sinx$ Then $\left(x^2+2x\right)\left(sinx\right)-\int \:\left(2x+2\right)\left(sinx\right)dx$ Now we integrate $\int\left(2x+2\right)\left(sinx\right)$ $u=2x+2$, $dv=sinx$ $du=2$, $v=-cosx$ So $\left(x^2+2x\right)\left(sinx\right)-\left[-\left(2x+2\right)cosx+2sinx\right]+C$ $\left(x^2+2x\right)\left(sinx\right)+\left(2x+2\right)cosx-2sinx+C$
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