Question

Use integration in parts to integrate [e^x(sinx)]dx. I made u=sin(x), du=cos(x)dx, v=e^x and dv=e^xdx. I solved for the whole thing and everything cancelled out, leaving me with C. Is this correct?

Answer 1

so you're saying you got

\[\Large \int e^x\sin(x)dx = C\]

??

Answer 2

if so, then it is incorrect

Answer 3

Bah! I had a feeling it was wrong.

Answer 4

That is the right equation though, yes.

Answer 5

is this one of those where you go around in a circle and get the same thing back again ? then divide by 2?

Answer 6

hopefully you have this so far?
\[\Large \int udv = uv - \int v du\]
\[\Large \int e^x\sin(x)dx = e^x\sin(x) - \int e^x \cos(x)dx\]

Answer 7

Yeah, that's what I got halfway through. Am I supposed to stop there or did I integrate wrong?

Answer 8

you have to use one more iteration of integration by parts

Answer 9

Ooooh. OH. I should've realized. WELP, let's see if I can make this work!

Answer 10

u = sin(x), du = cos(x)dx
dv = e^x, v = e^x


\[\large \int udv = uv - \int v du\]
\[\large \int e^x\sin(x)dx = e^x\sin(x) - \int e^x \cos(x)dx\]
\[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{\int e^x \cos(x)dx}\]


w = cos(x), dw = -sin(x)dx
dz = e^x, z = e^x

\[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[wz - \int z dw]}\]
\[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[e^x\cos(x) - \int e^x(-\sin(x)dx)]}\]
\[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{black}{[e^x\cos(x) + \int e^x\sin(x)dx]}\]
\[\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx\]

Answer 11

Yep, just worked out that part! And when I solve out for the last part of the equation, will I end up with e^xsin(x) - 2(e^xcos(x))?

Answer 12

very close

Answer 13

but no

Answer 14

Ah, wait, that would just cancel out again.

Answer 15

Let \[M = \int e^x\sin(x)dx\]

\[\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx\]
\[\large M = e^x\sin(x) - e^x\cos(x) - M\]

solve for M and re-substitute

Answer 16

Is it e^xsin(x) - e^xcos(x) + C?

Answer 17

nope

Answer 18

A right Ouroboros of pain and trig, this problem. Okay. What next?

Answer 19

Let \[M = \int e^x\sin(x)dx\]

\[\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx\]
\[\large M = e^x\sin(x) - e^x\cos(x) - M\]
\[\large M+M = e^x\sin(x) - e^x\cos(x) - M+M\]
\[\large 2M = e^x\sin(x) - e^x\cos(x)\]
\[\large M = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C\]
\[\large \int e^x\sin(x)dx = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C\]

Answer 20

Bless you, kind sir.

Answer 21

no problem