How do you integrate #int x^2 cos^2 3 x dx # using integration by parts?

1 Answer
Feb 29, 2016

#x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6#

Please check calculation too

Explanation:

#u = x^2#
#cos^2 3x dx= {1 + cos(6x)}/2 dx= dv#

#du = 2x dx#
#v = x/2 - sin(6x)/12#

integral #I = x^2(x/2-sin(6x)/12) -x^3/3 + int (x/6 sin(6x))dx#

integrate by parts again
#u=x#
#du=dx#
#dv = sin(6x)/6 dx#
#v= -cos(6x)#

so #I = x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6#