How do you integrate #int xsin(10x)# by integration by parts method?

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Answer 1 · 2017-01-29T08:35:37

#x/10cos10x+1/100sin10x+C#

It is important that the integration by parts formula be memorised

#intuv'dx=uv-intvu'dx#

when using the IBP the choice of #u " & " v'# is crucial.

in this case

#u=x=>u'=1#

#v'=sin10x=>v=-1/10cos10x#

#:.I=intuv'dx=uv-intvu'dx#

becomes

#I=-x/10cos10x-(int1/10cos10xdx)#

#I=-x/10cos10x+1/100sin10x+C#