Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises: 7

Answer

$$\int (x^2+2x)\cos xdx=(x^2+2x)\sin x+2\cos x(x+1)-2\sin x+C$$

Work Step by Step

$$A=\int (x^2+2x)\cos xdx$$ We would choose $u=x^2+2x$ and $dv=\cos xdx$ For $u=x^2+2x$, $du=(2x+2)dx=2(x+1)dx$ For $dv=\cos xdx$, $v=\sin x$ Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=(x^2+2x)\sin x-\int\sin x[2(x+1)dx]$$ $$A=(x^2+2x)\sin x-2\int(x+1)\sin xdx$$ We now choose $u'=x+1$ and $dv'=\sin xdx$ For $u'=x+1$, $du'=dx$. For $dv'=\sin xdx$, $v'=-\cos x$ Apply Integration by Parts a second time to A, we have $$A=(x^2+2x)\sin x-2[u'v'-\int v'du']$$ $$A=(x^2+2x)\sin x-2[-\cos x(x+1)-\int-\cos xdx]$$ $$A=(x^2+2x)\sin x-2[-\cos x(x+1)+\int\cos xdx]$$ $$A=(x^2+2x)\sin x-2[-\cos x(x+1)+\sin x+C]$$ $$A=(x^2+2x)\sin x+2\cos x(x+1)-2\sin x+C$$
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.