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 $intudv = uv - intvdu$

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

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

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

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

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