How do you integrate #int xsinxcosx# by integration by parts method?

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Answer 1 · 2016-11-28T17:46:19

The answer is #=(sin2x)/8-(xsin2x)/4+C#

We use
#sin2x=2sinxcosx#

#intxsinxcosxdx=1/2intxsin2xdx#

#intuv'=uv-intu'v#

#u=x#, #=>#, #u'=1#

#v'=sin2x#, #=>#, #v=-(cos2x)/2#

so, #intxsin2xdx=-(xcos2x)/2+1/2intcos2xdx#

#=-(xcos2x)/2+1/2*(sin2x)/2#

#=(sin2x)/4-(xcos2x)/2#

And finally

#intxsinxcosxdx=1/2((sin2x)/4-(xcos2x)/2) +C#

#=(sin2x)/8-(xsin2x)/4+C#