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

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

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

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