How do you integrate #int 2x sin 4x dx#?

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Answer 1 · 2016-01-20T09:44:17

#int2xsin(4x)dx = 1/8sin(4x)-1/2xcos(4x)+C#

Let #u = x# and #dv = sin(4x)dx#
Then #du = dx# and #v = -1/4cos(4x)#

So, by the integration by parts formula #intudv = uv - intvdu#

#2intxsin(4x)dx = 2(-1/4xcos(4x)-int((-1/4)cos(4x)dx)#

#=-1/2(xcos(4x)-intcos(4x)dx)#

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

#= 1/8sin(4x)-1/2xcos(4x)+C#