How do you integrate #x * cos^2 (x)#?

1 Answer
Rajat Chatterjee
Nov 7, 2016

#x^2/4+(x/4)sin2x+cos(2x)/8+c#.where c is integration constant .

Explanation:

#int (xcos^2x)dx#
#=(1/2)int(x2cos^2x)dx#
now #int(x2cos^2x)dx#
=#int{x(1+cos2x)}dx#
=#intxdx+int(xcos2x)dx#
[integration by parts]
=#x^2/2+x(int(cos2x)dx)-int[(dx/dx)int(cos2x)dx]dx#
=#x^2/2+(xsin(2x))/2##-int((sin2x)/2)dx#
=#x^2/2+(xsin(2x))/2##+(cos2x)/4##+c#
hence the value of the integral :#x^2/4+(xsin(2x))/4+cos(2x)/8+c#,where c is constant of integration.

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