How do you integrate #int xcos(2x)# by integration by parts method?

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Answer 1 · 2016-08-19T13:59:49

#1/4(2x*sin2x+cos2x)+C#.

Let #I=intxcos2xdx#

We use the Rule of Integration by Parts, which is,

#intuvdx=uintvdx-int[(du)/dxintvdx]dx#

We take,

#u=x, so (du)/dx=1; & v=cos2x, so, intvdx=(sin2x)/2#.

Hence, #I=x/2*sin2x-1/2intsin2xdx#

#=x/2*sin2x-1/2((-cos2x)/2)#

#=1/4(2x*sin2x+cos2x)+C#.

Enjoy Maths.!