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

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Answer 1 · 2016-08-14T09:25:06

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

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

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

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