integration by parts

[nickname]

?xsinxdx

using ?udv=uv-?vdu

I thought that for this problem x was going to be u and sinx dv,

but the solution is the following

?xsinxdx x is dv and u is sinx

= x(-cosx) - ?(-cosx) 1dx
= -xcosx+ ?(cosx) dx
= -xcosx+sinx+c

this is the same as sinx-xcosx+c

could anyone please clarify why x in dv instead of u? I thought that that the simplest term should be used as u.

Thanks again!

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \int x sin(x)dx, \ I \ by \ P: \ \int udv \ = \ uv \ -\int v du\)

\(\displaystyle Let \ u \ = \ x \ \implies \ du \ = \ dx, \ and \ let \ dv \ = \ sin(x)dx \ \implies \ v \ = \ -cos(x)\)

\(\displaystyle Ergo, \ we \ have \ \int x sin(x)dx \ = \ -xcos(x) \ + \ \int cos(x)dx\)

\(\displaystyle \int x sin(x)dx \ = \ -xcos(x) \ + \ sin(x) \ + \ C, \ QED\)

\(\displaystyle Note, \ above \ comment, \ to \ wit:\)

\(\displaystyle could \ anyone \ please \ clarify \ why \ x \ in \ dv \ instead \ of \ u? \ I \ thought \ that \ the\)

\(\displaystyle \ simplest \ term \ should \ be \ used \ as \ u.\)

\(\displaystyle You \ thought \ wrong, \ however; \ if \ you \ want \ to \ make \ it \ hard \ upon \ yourself, \ go \ for \ it.\)