This integral can be easily solved using a consequence of Moivre's identity
(e^{ix}+e^{-ix})/2 = cos(x)eix+e−ix2=cos(x)
(e^{ix}-e^{-ix})/(2i) = sin(x)eix−e−ix2i=sin(x)
because
cos^2x equiv (e^{2ix} +2+e^{-2ix})/4cos2x≡e2ix+2+e−2ix4
then
inte^(3x)cos^2xdx equiv 1/4int (e^{3x+2ix} +2e^{3x}+e^{3x-2ix})dx∫e3xcos2xdx≡14∫(e3x+2ix+2e3x+e3x−2ix)dx
=e^{3x}/4((e^{2ix})/(3+2i)+2/3+e^{-2ix}/(3-2i))=e3x4(e2ix3+2i+23+e−2ix3−2i)
=e^{3x}/4((3-2i)/13e^{2ix}+2/3+(3+2i)/13e^{-2ix})=e3x4(3−2i13e2ix+23+3+2i13e−2ix)
=e^{3x}(3/26(e^{2ix}+e^{-2ix})/2+1/13(e^{2ix}-e^{-2ix})/(2i)+1/6)=e3x(326e2ix+e−2ix2+113e2ix−e−2ix2i+16)
=e^{3x}/78(9cos(2x)+6sin(2x)+13)=e3x78(9cos(2x)+6sin(2x)+13)
