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.!

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