## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises: 31

#### Answer

$= \pi$

#### Work Step by Step

$\begin{gathered} \int_0^\pi {x\sin x\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = dx \hfill \\ dv = \sin xdx\,\,\,\, \to \,\,\,v = - \cos x \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = - x\cos x - \int_{}^{} {\,\left( { - \cos x} \right)dx} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ - x\cos x + \sin x + C \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \int_0^\pi {x\sin x\,dx} = \,\,\left[ { - x\cos x + \sin x} \right]_0^\pi \hfill \\ \hfill \\ evaluate\,\,the\,\,{\text{limits}} \hfill \\ \hfill \\ = \,\left( { - \pi \cos \pi + \sin \pi } \right) - \,\left( 0 \right) = \pi \hfill \\ \end{gathered}$

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