How do you find the integral of #x cos 2x dx#?

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Aug 30, 2015

Answer:

#int xcos2xdx=x/2sin2x +1/4cos2x +c#

Explanation:

Use integration by parts

#int xcos2xdx= x intcos2xdx - int (d/dx (x)intcos2xdx)dx#

#= x 1/2sin2x - int 1/2sin2xdx#

#= x/2sin2x -1/2 (-1/2cos2x)+c#

#= x/2sin2x +1/4cos2x +c#

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