Integration by parts query. Getting 0 as answer, think its wrong, help!!

Hi,

I'm asked to integrate e^2xcosxdx. I have used intefration by parts letting u=e^2x and dv=cosxdx. This results in

e^2xsinx-the integral of 2sinxe^2x. I then integrated by parts again and the overall result was if I = integral of e^2xcosxdx then

I= e^2xsinxdx-e^2xsinxdx + 0.5( I)

0.5I = 0

Therefore I=0

Is this wrong?

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

Quote:

Originally Posted by

**orlacoon** Hi,

I'm asked to integrate e^2xcosxdx. I have used intefration by parts letting u=e^2x and dv=cosxdx. This results in

e^2xsinx-the integral of 2sinxe^2x. I then integrated by parts again and the overall result was if -I = integral of e^2xcosxdx then

I= e^2xsinxdx-e^2xsinxdx + 0.5( I)

0.5I = 0Therefore I=0

Is this wrong?

You are making a sign error.

Add that last integral to both sides and then divide by 2.

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

Still not sure where I'm going wrong. Could you explain further? Thanks

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

You should end up with the same integration but with a minus sign on the right side. You move that integration to the left hand side and then divide by two.

So you should end up with something like:

I= .......... -I (move right hand integration to left hand side)

2I = .............. (divide by two)

I = 1/2(..........) (Final answer)

Sorry i didn't do the integration. In class right now =p

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

Quote:

Originally Posted by

**Plato** You are making a sign error.

Add that last integral to both sides and then divide by 2.

He's right. Remember to pick the right u and v' elements.

Derivative of (sinx)' = cosx and (cosx)' = -sinx

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

$\displaystyle \int e^{2x}\cos{x} \, dx$

$\displaystyle u = \cos{x} , du = -\sin{x} \, dx$

$\displaystyle dv = e^{2x} \, dx , v = \frac{1}{2}e^{2x}$

$\displaystyle \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int e^{2x} \sin{x} \, dx$

$\displaystyle p = \sin{x} , dp = \cos{x} \, dx$

$\displaystyle dq = e^{2x} \, dx , q = \frac{1}{2}e^{2x}$

$\displaystyle \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\left[\frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\int e^{2x}\cos{x} \, dx\right]$

$\displaystyle \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{4}e^{2x}\sin{x} - \frac{1}{4}\int e^{2x}\cos{x} \, dx$

$\displaystyle \frac{5}{4} \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{4}e^{2x}\sin{x} + C$

$\displaystyle \int e^{2x}\cos{x} \, dx = \frac{2}{5}e^{2x}\cos{x} + \frac{1}{5}e^{2x}\sin{x} + C$

Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

Thanks a lot, got it wrong way around!