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

1 Answer
Cesareo R.
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)

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