How do you integrate #inte^(3x)cos^2xdx# using integration by parts?

1 Answer
Jul 22, 2016

#inte^(3x)cos^2xdx=e^{3x}/78(9cos(2x)+6sin(2x)+13)#

Explanation:

This integral can be easily solved using a consequence of Moivre's identity

#(e^{ix}+e^{-ix})/2 = cos(x)#
#(e^{ix}-e^{-ix})/(2i) = sin(x)#

because

#cos^2x equiv (e^{2ix} +2+e^{-2ix})/4#

then

#inte^(3x)cos^2xdx equiv 1/4int (e^{3x+2ix} +2e^{3x}+e^{3x-2ix})dx#

#=e^{3x}/4((e^{2ix})/(3+2i)+2/3+e^{-2ix}/(3-2i))#
#=e^{3x}/4((3-2i)/13e^{2ix}+2/3+(3+2i)/13e^{-2ix})#
#=e^{3x}(3/26(e^{2ix}+e^{-2ix})/2+1/13(e^{2ix}-e^{-2ix})/(2i)+1/6)#
#=e^{3x}/78(9cos(2x)+6sin(2x)+13)#