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How do you integrate #int tsin(2t)# by integration by parts method?

1 Answer

Aug 14, 2016

#I = -t cos(2t)/2 + 1/4 sin(2t)#

Explanation:

Set #sin(2t)dt = dv#
then #v = -cos(2t)/2#
#u = t#
#du = dt#

Using #int udv = [uv] - int v du#
now you get
#I = -t cos(2t)/2 + 1/2 int cos(2t) dt#
#I = -t cos(2t)/2 + 1/4 sin(2t)#

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