Integration using Eulers formual

[cragar]

Homework Statement

e^(-2x)*sin(3x)dx

Homework Equations

using e^(ix)=cosx+isinx

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.

 

[Mark44]

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

 

[cragar]

cause i think e^(3ix)=cos(3x)+isin(3x) i don't know i know it works with
e^(-2x)*(cos(3x)

 

[rock.freak667]

cause i think e^(3ix)=cos(3x)+isin(3x) i don't know i know it works with
e^(-2x)*(cos(3x)

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

 

[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.

 

[rock.freak667]

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.

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

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= $\int e^{2x}sin3x dx$

 

[Mark44]

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.

 

[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 .

 

[Mark44]

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

 

[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

 

[rock.freak667]

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

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.

 

[cragar]

k i think i get it now.