# Maths integration question

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#1
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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17 years ago
#2
(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.

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).

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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