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Integration by Parts

Hopefully somebody will be able to help me with this.
For some reason I am ending up with 0=0 for an answer, I think it may be me getting signs mixed up somewhere.
Here is my question and workings out:-
∫exSin2xdx \int e^xSin2xdx
I=∫exSin2xdxI=\int e^xSin2xdx
letu=exlet u=e^x
dudx=ex\frac {du}{dx}=e^x
dvdx\frac {dv}{dx}Sin2xSin2x
v=−12v=- \frac{1}{2}Cos2xCos2x
I=ex∗−12I=e^x*-\frac {1}{2}Cos2x−∫−12Cos2x - \int-\frac{1}{2}Cos2x∗exdxCos2x*e^xdx
I=−12I=-\frac{1}{2}exCos2x+∫12e^xCos2x+\int \frac{1}{2}exCos2xdxe^xCos2xdx
letu=Cos2xlet u=Cos2x
dudx\frac{du}{dx}=−2Sin2x=-2Sin2x
dvdx\frac{dv}{dx}=12=\frac{1}{2}exe^x
v=12v=\frac{1}{2}exe^x
I=−12exCos2x+12exCos2x−∫12ex−2Sin2xdxI=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xCos2x-\int\frac{1}{2}e^x-2Sin2xdx
I=−12exCos2x+12exCos2x−∫−exSin2xdxI=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xCos2x-\int-e^xSin2xdx
I=−12exCos2x+12exCos2x+∫exSin2xdxI=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xCos2x+\int e^xSin2xdx
I=−12exCos2x+12exCos2x+II=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xCos2x+I
I+I=−12exCos2x+12exCos2xI+I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xCos2x
0=00=0

Where am I going wrong?

Barney
Reply 1
Avatar for Hasufel
Hasufel
16
you have to stick to using the same functions for u and dv/dx, otherwise...

(i.e. for the 2nd integration, stick with u=exp term, dv/dx = trig term)
For the second by parts let u=12ex u = \frac{1}{2}e^x again and dvdx=cos2x\frac{dv}{dx} = cos2x

Keep the u and dv/dx the same parts.
Reply 3
Avatar for Barney63
Barney63
OP
So using:-
u=12exu=\frac{1}{2}e^x
dudx=12ex\frac{du}{dx}=\frac{1}{2}e^x
dvdx=Cos2x\frac{dv}{dx}=Cos2x
v=SinxCosxv=SinxCosx
I get:-
I=−12exCos2x+12exSinxCosx−∫12exSinxCosxdxI=-\frac{1}{2}e^xCos2x+\frac{1}{2}e^xSinxCosx-\int \frac{1}{2}e^xSinxCosxdx
Where do I get the ∫exSin2xdx\int e^xSin2xdx for the substitution of I?

Barney
Reply 4
Avatar for Hasufel
Hasufel
16
if
Unparseable latex formula:

\diasplaystyle \frac{dv}{dx}=cos(2x)

the,

∫cos(2x)dx=12sin(2x) \int cos(2x)dx=\frac{1}{2}sin(2x)

giving the last part as −14∫exsin(2x)\frac{-1}{4}\int e^{x}sin(2x)

then - because you have a multiple of the same integral on the LHS (muliply both sides by 4 and you`ll see this), you can put the same one all on one side of the equals sign and...
(edited 11 years ago)
Reply 5
Avatar for Barney63
Barney63
OP
I thought you only did the derivative of u, and that you integrated dvdx\frac{dv}{dx}.

Barney
Reply 6
Avatar for Hasufel
Hasufel
16
But, i was pointing out that, when you integrated your dv/dx you got Sin(x)Cos(x), when in fact the answer`s (1/2)Sin(2x).

Hang on.. i`ll type out what you should get (but not the answer, `cause that`s the easy part - you`ve basically got it, but i think the arithmetic is maybe freaking you out!)
(edited 11 years ago)
Reply 7
Avatar for Hasufel
Hasufel
16
∫e2sin(2x)dx=−12cos(2x)−−12∫excos(2x)dx\int e^{2}sin(2x)dx=\frac{-1}{2}cos(2x)- \frac{-1}{2} \int e^{x}cos(2x)dx

taking the second part:

+12∫excos(2x)dx=12(12exsin(2x)−12∫exsin(2x)dx)\frac{+1}{2} \int e^{x}cos(2x)dx= \frac{1}{2}(\frac{1}{2}e^{x}sin(2x)-\frac{1}{2} \int e^{x}sin(2x)dx)

giving, together:

∫e2sin(2x)dx=−12cos(2x)+14exsin(2x)−14∫exsin(2x)dx\int e^{2}sin(2x)dx=\frac{-1}{2}cos(2x)+\frac{1}{4}e^{x}sin(2x)-\frac{1}{4} \int e^{x}sin(2x)dx

multiply this by 4, and take the last term over to the LHS...
(edited 11 years ago)
Reply 8
Avatar for Barney63
Barney63
OP
Sorry for the delay and also potentially being thick, but where did the ∫e2Sin(2x)dx\int e^2Sin(2x)dx come from?

Barney
Reply 9
Avatar for Barney63
Barney63
OP
!!!!Bump!!!!
Original post by Barney63
Sorry for the delay and also potentially being thick, but where did the ∫e2Sin(2x)dx\int e^2Sin(2x)dx come from?

Barney

From e22sinxcosxe^22sinxcosx
Reply 11
Avatar for Hasufel
Hasufel
16
maybe this will explain it better...

http://www.youtube.com/watch?v=1bSTXcFeaZ4

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