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## Main Question or Discussion Point

i'm trying to solve an equation, but i'm stuck on this step:

(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?

(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?

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i'm trying to solve an equation, but i'm stuck on this step:

(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?

(integral sign) e^2x*e^x(3sin2x+2cos2x)dx =?

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[tex]\int e^{(a+ib)x}dx[/tex]

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arildno

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Suppose you are to find an anti-derivative (i.e, indefinite integral) of the function [tex]f(x)=e^{x}\sin(x)[/tex], that is, you are to find J, where J is given as:

[tex]J=\int{e}^{x}\sin(x)dx(1)[/tex]

The right-hand side can now be rewritten as:

[tex]\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-\int{e}^{x}\cos(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-\int{e}^{x}\sin(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-J(2)[/tex]

where I have used integration by parts twice, along with (1).

Thus, we have:

[tex]J=e^{x}\sin(x)-e^{x}\cos(x)-J\to{J}=\frac{e^{x}\sin(x)-e^{x}\cos(x)}{2}[/tex]

(I've not bothered with the constant of integration; this should also be included in the final expression).

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ok, thanks! i'll try that~

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come to think of it... is there a quicker way?

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arildno

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haha... thanks! :)

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