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

Question

3740 views around the world

You can reuse this answer
Creative Commons License

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:

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

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

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

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

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

How do you integrate int xsin2xint xsin2x by parts from [0,pi][0,pi]?