How do you integrate #int xsin2x# by parts from #[0,pi]#?

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Answer 1 · 2017-02-05T13:59:58

#int_0^pi xsin2x dx =-pi/2#

Consider the integral:

#int_0^pi xsin2x dx#

Note that:

#d(-1/2cos2x) = sin2x dx#

so we can integrate by parts:

#int_0^pi xsin2x dx = int_0^pi xd(-1/2cos2x)#

#int_0^pi xsin2x dx = [-(xcos2x)/2]_0^pi +1/2 int _0^pi cos2x dx#

#int_0^pi xsin2x dx = [-(xcos2x)/2]_0^pi +1/4 [sin2x]_0^pi #

#int_0^pi xsin2x dx = [-(picos(2pi))/2 +0] + [0 -0]#

#int_0^pi xsin2x dx =-pi/2#