Integrate e^x with me (1 Viewer)

[conics2008]

find S e^(1+i)x ???

USING



e^a+ib = e^a ( cosb + isinb)

which gives us e^(1+i)x = e^x+ix = e^x ( cosx+isinx) = e^xcosx+ie^xsinx

now apply IBP on that.. the prob is S ie^xsinx = 0 ???

 

[independantz]

 

[conics2008]

thats wrong independantz, you cant just do that =[

i got up to e^x/2 (cosx+sinx) ??

but the book doesn't have any solutions.

i is imaginary ?

 

[tommykins]

 

[conics2008]

tommy yeh done that ?? but when u get to this e^xisinx using integration by parts it stuffs everything up =[

 

[conics2008]

I got it =]

simple sign mistake
€
can some one confirm this solution please.
is it

e^x/2 (cosx+sinx) + ie^x/2(sinx-cosx) = e^x/2( cosx+sinx +i(sinx-cosx) ) + C ???

 

[tommykins]

post solution plz

 

[conics2008]

The question was find S e^(1+i)x dx

using the fact e^iy = cosy+isiny

therefore

e^x+ix = e^x * e^ix = e^x(cosx+isinx)

therefore S e^xcosx +ie^xsinx = Se^(1+i)x

Using integration by parts on the expanded version

S e^xcosx dx = e^x/2 (cosx+sinx)

and S e^xisinx dx = ie^x/2 (sinx-cosx)

just add them to and simplify =]

 

[lolokay]

this is a weird question, for asking you to do the integral in a more complicated way than necessary

where is the question from?

 

[lolokay]

so you can't even assume that

?

 

[azureus88]

im very sure that you can take "i" as a constant since this assumption was used to prove e^i@ = cis@.

 

[independantz]

independantz is no more incorrect than u conics2008, as u both are essentially assuming i can be treated like a constant. essentially, ur answer and independantz's answers are identical, its just that u used more convoluted methods to get to it:

But isnt "i" a constant?, from what I gather it is always equivalent to

 

[lolokay]

so how are you supposed to do this question?

 

[shaon0]

thats wrong independantz, you cant just do that =[

i got up to e^x/2 (cosx+sinx) ??

but the book doesn't have any solutions.

i is imaginary ?

Dude, it's the same thing. Either method is correct.

 

[conics2008]

guys read up, i posted the solution, its the same as the other guy whos tellin me its wrong ??

go read up where it says the + C ???

 

[conics2008]

independantz is no more incorrect than u conics2008, as u both are essentially assuming i can be treated like a constant. essentially, ur answer and independantz's answers are identical, its just that u used more convoluted methods to get to it:

€


hey if you read up, i got the same thing as you, and i treated i as a constant because it came from square root of -1 ??

 

[Js^-1]

I got this ^
Is that the correct answer? Also, does the statement

hold for complex a?

 

[Timothy.Siu]

€


hey if you read up, i got the same thing as you, and i treated i as a constant because it came from square root of -1 ??

that was his point, that its the same.

 

[Templar]

I got this ^
Is that the correct answer? Also, does the statement

hold for complex a?

Yes, it holds for complex a.

You would have considerable problems if you tried to integrate between two values in the complex though.

 

[shaon0]

Yes, it holds for complex a.

You would have considerable problems if you tried to integrate between two values in the complex though.

So i could not integrate:
S [limits: 2i to i] x dx
= 0.5(2i)^2-0.5(i)^2
= -2+0.5
= -1.5