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

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Answer 1 · 2016-06-22T22:25:44

#1/2 e^x (cos x + sin x) + C #

#int \ e^x cos x \ dx#

#= \mathcal{Re} ( int \ e^x (cos x + i \ sin x) \ dx ) #

#= \mathcal{Re} ( int \ e^x e^{i x} \ dx) #

#= \mathcal{Re} ( int \ e^{(1+i)x} \ dx) #

#= \mathcal{Re} ( 1/(1 + i ) e^{(1+i)x} ) #

#= e^x \mathcal{Re} ( 1/(1 + i ) e^{ix} ) #

#= e^x \mathcal{Re} ( (1-i)/((1 + i)(1-i) ) (cos x + i sin x) ) #

#= e^x \mathcal{Re} ( (1-i)/(2) (cos x + i sin x) ) #

#=1/2 e^x \mathcal{Re} ( (1-i) (cos x + i sin x) ) #

#=1/2 e^x \mathcal{Re} (cos x + i sin x - i cos x + sin x ) #

#=1/2 e^x \mathcal{Re} ((cos x + sin x) + i( sin x - cos x) ) #

#=1/2 e^x (cos x + sin x) + C #