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iiikewldude
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By integratiing by parts twice, show that :-

(integral) f(x) dx = (1/2)*e^x (Cos x + Sin x) + c

where f(x) = e^x Cos x

Cheers to anyone who can do this. I kind of seem to get stuck in a loop when doing it. If you try it you might find out what i mean.
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Jonny W
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Report 17 years ago
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(Original post by iiikewldude)
By integratiing by parts twice, show that :-

(integral) f(x) dx = (1/2)*e^x (Cos x + Sin x) + c

where f(x) = e^x Cos x

Cheers to anyone who can do this. I kind of seem to get stuck in a loop when doing it. If you try it you might find out what i mean.
Let

I(y) = (int from 0 to y) e^x cos(x) dx,
J(y) = (int from 0 to y) e^x sin(x) dx.

(It's neater to work with definite integrals, but we can switch to indefinite integrals at the end.) We work out I(y) and J(y) by finding two equations relating them.

First, start with I(y) and integrate by parts:

I(y)
= [ e^x sin(x) ](from 0 to y) - (int from 0 to y) e^x sin(x) dx
= e^y sin(y) - J(y).

Second, start with J(y) and integrate by parts:

J(y)
= [ -e^x cos(x) ](from 0 to y) + (int from 0 to y) e^x cos(x) dx
= 1 - e^y cos(y) + I(y).

So, eliminating J(y),

I(y) = e^y sin(y) - 1 + e^y cos(y) - I(y).

So,

I(y) = (1/2)*(e^y sin(y) + e^y cos(y) - 1).

For the indefinite integral we can forget the -1/2, and so

(int) e^x cos(x) dx = (1/2)*(e^x sin(x) + e^x cos(x)) + c.
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