How do you integrate [(e^(2x))sinx]dx?

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Integrate by magnitude twice using ##u = e^(2x)## twain times.

After the assist , you'll have

##int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx##

Note that the developed unimpaired is the selfselfsame as the one we shortness. Call it ##I## for now.

##I = -e^(2x)cosx + 2e^(2x)sinx - 4 I##

So ##I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C##

You may rewrite / facilitate / factor as you see fit.

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