How do you integrate $int xe^xsinx$ by integration by parts method?

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Answer 1 · 2016-09-14T09:19:50

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

$int xe^xsinx dx$

$= mathcal{I}{ int xe^xe^(ix) dx }$

$= mathcal{I}{color(red)( int xe^((1+i)x)dx) }$

working the bit in red by IBP

$int xe^((1+i)x)dx$

$= int xd/dx (1/(1+i) e^((1+i)x)) dx$

$= x/(1+i) e^((1+i)x) - int d/dx (x) 1/(1+i) e^((1+i)x) dx + C$

$= x/(1+i) e^((1+i)x) - int 1/(1+i) e^((1+i)x) dx + C$

$= x/(1+i) e^((1+i)x) - 1/(1+i)^2 e^((1+i)x) + C$

$= (1-i)/(1-i) x/(1+i) e^((1+i)x) - ((1-i)/(1-i))^2 1/(1+i)^2 e^((1+i)x) + C$

$= (1-i) x/2 e^((1+i)x) - (1-i)^2 1/4 e^((1+i)x) + C$

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

And so we want

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

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

How do you integrate int xe^xsinxint xe^xsinx by integration by parts method?