Integration by parts

[thatstheguy9]

[MATH] \int e^xsin(x) dx [/MATH]
For this problem I chose to let:
[MATH] u = sin(x), du = cos(x)dx [/MATH] AND
[MATH] dv = e^x dx, v = e^x [/MATH]
Therefore
[MATH]\int sin(x)e^x = sin(x)e^x - \int e^x cos(x) dx[/MATH][MATH] \int sin(x)e^x = sin(x)e^x -e^xcos(x) [/MATH][MATH]\int sin(x)e^x = e^x(sin(x) - cos(x)) + C [/MATH]
Other solutions choose to let u = e^x and vice versa to find the correct result.
Can someone explain why letting u = sin(x) is a poor choice and leads to the incorrect solution?

 

[Romsek]

Explain the transition between the 1st and 2nd lines after the word Therefore.

You seem to just ditch an integral sign with no explanation.

 

[thatstheguy9]

Explain the transition between the 1st and 2nd lines after the word Therefore.

You seem to just ditch an integral sign with no explanation.

It seems I was only taking the integral of v, i.e [MATH] \int sin(x)e^x = sin(x)e^x - "\int e^x"[COLOR=rgb(0, 0, 0)] [/COLOR]cos(x) dx [/MATH]and not the entire term [MATH] \int e^x cos(x) dx [/MATH]
My guess is taking the integral of the entire term would result in another integration by parts, and so on. Which would explain why choosing those terms for u and dv doesn't work out?

 

[Romsek]

If I recall you have to apply integration by parts twice. And then you can do some algebra and group common terms
that gets you a nice closed form solution.

 

[thatstheguy9]

If I recall you have to apply integration by parts twice. And then you can do some algebra and group common terms
that gets you a nice closed form solution.

I see, thanks for the help mate

 

[Steven G]

Romsek is 100% correct. You did make the correct choice for u. Do integration by parts again, letting u = cosx this time. You will then end up with a multiple of the original integral. Combine this multiple of the original integral with the one from the left hand side. The results will follow after a simple division.

Post back showing your work.

If I recall correctly, you could have chosen u=e^x both times and that would work out to the same results.