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How do you integrate #int xsin(2x)# by integration by parts method?

1 Answer

Aug 8, 2016

#=1/4sin(2x) - x/2cos(2x) + C#

Explanation:

For #u(x), v(x)#

#int uv'dx = uv ' - int u'vdx#

#u(x) = x implies u'(x) = 1#

#v'(x) = sin(2x) implies v(x) = -1/2cos(2x)#

#intxsin(2x)dx = -x/2cos(2x) +1/2intcos(2x)dx#

#= -x/2cos(2x)+1/4sin(2x)+C#

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