What is the integral of # x * cos^2 (x)#?

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Answer 1 · 2016-06-30T09:10:51

#= 1/4 x sin 2x + 1/8 cos 2x + x^2/4+ C#

#int dx qquad x * cos^2 (x)#

it will be easier first to use the double angle formula #cos 2A = 2 cos^2 A - 1# or #cos^2 A = (cos 2A + 1)/2#

so we are looking at

#1/2 int dx qquad color{red}{x cos 2x} + x#

as it's a composite term, we should do the red bit using IBP ie

#int u v' = uv - int u' v#

here

#u = x, u' = 1#
#v' = cos 2x, v = 1/2 sin 2x#

so

#1/2 int dx qquad color{red}{x cos 2x} + x#

#= 1/2 { 1/2 x sin 2x - int dx qquad 1/2 sin 2x +int dx qquad x }#

#= 1/2 { 1/2 x sin 2x + 1/4 cos 2x + x^2/2 } + C#

#= 1/4 x sin 2x + 1/8 cos 2x + x^2/4+ C#