using by parts integrate cos(log x)dx

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Please find below the solution to the asked query:

I=coslogx.dx ;i=1.coslogx.dxUsing integration by part taking coslogx as first function and 1 as secondfunction.I=coslogx1.dx-ddxcoslogx1.dxdx=coslogx.x--sinlogx.1xx.dx  ddxcoslogx=-sinlogx.ddxlogx=-sinlogx.1x=coslogx.x+sinlogx.dxI=x.coslogx+1.sinlogx.dxAgain using integration by part taking sinlogx as first function and 1 as secondfunctionI=x.coslogx+sinlogx1.dx-ddxsinlogx1.dxdxI=x.coslogx+sinlogx.x-coslogx.1x.xdxI=x.coslogx+x.sinlogx-coslogxdxUsing i we have I=coslogx.dxI=x.coslogx+sinlogx.x-II+I=xcoslogx+sinlogx2I=xcoslogx+sinlogxI=x2coslogx+sinlogx+C, where C is integration constant.coslogx.dx=x2coslogx+sinlogx+C

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