# Integration using Eulers formual

1. Mar 14, 2009

### cragar

1. The problem statement, all variables and given/known data
e^(-2x)*sin(3x)dx

2. Relevant equations
using e^(ix)=cosx+isinx

3. The attempt at a solution
so i get e^(-2+i3)(x)

so then after i integrate i get (e^(-2+i3)(x))/(-2+i3) then i multiply by the conjegate to get the i out of the bottom then i multiply it by cos(3x)+isin(3x) then pick out the real part is this correct.

2. Mar 15, 2009

### Staff: Mentor

So how do you get from e-2xsin(3x) to e(-2 + 3i)x?

3. Mar 15, 2009

### cragar

cause i think e^(3ix)=cos(3x)+isin(3x) i dont know i know it works with
e^(-2x)*(cos(3x)

4. Mar 15, 2009

### rock.freak667

then you'd want to extract the imaginary component to get the required integral, right?

5. Mar 15, 2009

### cragar

yes but the sin(3x) is throwing me
so in the formula e^(ix)=cos(x)+isin(x) in that formula the cosx is the real part
but in our case we need the sinX to be the real part so do we need to manipluate the formula to make it work am i no the right track.

6. Mar 15, 2009

### rock.freak667

$$\int e^{2+3i}x dx = \int e^{2x}e^{3xi}dx=\int (e^{2x}cos3x+ie^{2x}sin3x)dx=A+Bi$$

A+Bi is what you realized when you integrated exp(2+3i)

$$\int (e^{2x}cos3x+ie^{2x}sin3x)dx = \int e^{2x}cos3x dx + i \int e^{2x}sin3x dx =A+iB$$

So if you equate the real and imaginary components...

the imaginary part B= $\int e^{2x}sin3x dx$

7. Mar 15, 2009

### Staff: Mentor

I believe you're on the wrong track. Another approach is to integrate by parts twice. After the first application the part you integrate will be something like e^(-2x)cos(3x). Apply integration by parts one more time to get an integral similar to the one you started with. You'll should have an equation in which you can solve algebraically for the integral you want.

8. Mar 15, 2009

### cragar

the whole reason for doing this is to avoid integration by parts which i know how to do it that way. i want to learn how to do it this way .

9. Mar 15, 2009

### Staff: Mentor

OK, then rock.freak's and your direction is the way to go...

10. Mar 15, 2009

### cragar

ok so would i then divide everything by i in the formula to make sinx the real part then use that so i would have e^(ix)/i=cosx/i + sinx or is that wrong

11. Mar 15, 2009

### rock.freak667

I still don't get why you want to make sin on the left side in the integral for.When you get the right side in the form A+Bi. B is supposed to be the required integral.

12. Mar 15, 2009

### cragar

k i think i get it now.