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

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

$int \ e^x cos x \ dx$

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

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

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

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

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

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

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

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

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

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

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

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