How do you integrate $int xsin2x$ by integration by parts method?

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Answer 1 · 2017-01-16T11:46:55

$int xsin2xdx = -1/2xcos2x +1/4sin2x$

As:

$d(cos2x) =-2sin2x dx$

we can integrate by parts in this way:

$int xsin2xdx = -1/2 int x d(cos2x) = -1/2 xcos2x +1/2 int cos2x dx= -1/2xcos2x +1/4sin2x$

How do you integrate int xsin2xint xsin2x by integration by parts method?