#How to find integral of this

19 messages · Page 1 of 1 (latest)

last sail
wind kelpBOT
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last sail
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So I’m suppose to move the right e^-xcos(2x) to the left but wouldn’t that make it 0 or something

hollow glade
# last sail

Ah, this is a famous one.
If you have an integral of e^(ax) cos(bx) (or e^(ax) sin(bx), doesn't matter), then after integrating by parts twice you get the same integral.
For example, for the cosine case:
∫(e^(ax) cos(bx)dx) = e^(ax) ((1/a)cos(bx) + (b/a^2)sin(bx)) - (b^2/a^2)∫(e^(ax) cos(bx)dx)
And the approach here is to take that integral as I. So:
I = e^(ax) ((1/a)cos(bx) + (b/a^2)sin(bx)) - (b^2/a^2)I
And now try expressing I from here.

fading seal
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Its 1/2 of it

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x-x/2=x/2

last sail
last sail
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So there is no more of that integral?idk

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So this becomes I-2I

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I got this

IMG_5193.jpg
fading seal
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Look.

midnight pewterBOT
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@last sail

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@fading seal @hollow glade The user still needs help with this help request.

last sail
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+close

midnight pewterBOT
# last sail +close
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