What is #int (((e^(ix)+(-e^(-ix))/(2))dx#?

1 Answer
Eddie
Jun 24, 2016

#= -3/2 i cos x + 1/2 sin x + D#

Explanation:

#int \ e^(ix)+(-e^(-ix))/(2) \ dx# = #int \ e^(ix)- 1/2 e^(-ix) \ dx#

#= 1/i e^(ix)-1/(-i) (e^(-ix))/(2) + C#

#= -i e^(ix) - i (e^(-ix))/(2) + D#

#= -i (cos x + i sin x) -i (cos x - i sin x)/(2) + D#

#= -i cos x + sin x - i/2 cos x - i/2 sin x + D#

#= -3/2 i cos x + 1/2 sin x + D#

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