Question

I'm having trouble integrating the following by parts:

S(e^(-x)*sin(4x)dx

Whenever I try to integrate I end up in this infinite loop situation. How do I get around this?

Answer 1

\[\int\limits_{}^{}e^{-x}\sin(4x)dx\] To be clear, that's my integral.

Answer 2

Keep integrating by parts, eventually it will oscillate back to:
\[\int\limits_{}^{} e ^{-x}\sin(4x) dx\]
and you just subtract both sides by that^

Answer 3

Oh, that seems so simple, I guess I kept thinking I'd have to just go down an endless rabbit hole, but once it's on both sides I can subtract it off. Haha thanks.

Answer 4

Let:

u = e^-x

du = (-e^-x) dx

dv = Sin(4x) dx

v = -Cos(4x) / 4

so:

\[\int\limits_{}^{} e ^{-x}\sin(4x) dx = -e ^{-x}(1/4)Cos(4x) - \int\limits_{}^{} -e ^{-x} (-Cos(4x)(1/4))dx\]

\[\int\limits_{}^{} e ^{-x}\sin(4x) dx = -e ^{-x}Cos(4x) -(1/4)\int\limits_{}^{} e ^{-x}Cos(4x)dx\]

Now integrate by parts again:

let u = e^-x

du = -e^-x dx

dv = Cos(4x)dx

v = (1/4) Sin(4x)

\[\int\limits\limits_{}^{} e ^{-x}\sin(4x) dx = -e ^{-x}Cos(4x) -(1/4)[e ^{-x}(1/4)Sin(4x)- \int\limits_{}^{}-e ^{-x}Sin(4x)(1/4)dx]\]
\[\int\limits\limits\limits_{}^{} e ^{-x}\sin(4x) dx = -e ^{-x}Cos(4x) -(1/16)e ^{-x}Sin(4x)+(1/16) \int\limits\limits_{}^{}e ^{-x}Sin(4x)dx\]

Do you see the:
\[(1/16) \int\limits\limits_{}^{}e ^{-x}Sin(4x)dx\]

you see how it shows up on both sides of the equation? lets bring it to one side:

\[\int\limits\limits\limits\limits_{}^{} e ^{-x}\sin(4x) dx -(1/16) \int\limits\limits\limits_{}^{}e ^{-x}Sin(4x)dx = -e ^{-x}Cos(4x) -(1/16)e ^{-x}Sin(4x)\]

\[(15/16)\int\limits\limits\limits\limits_{}^{} e ^{-x}\sin(4x) dx = -e ^{-x}Cos(4x) -(1/16)e ^{-x}Sin(4x)\]

now just bring the (15/16) back to the other side:
\[\int\limits\limits\limits\limits_{}^{} e ^{-x}\sin(4x) dx = (16/15) [-e ^{-x}Cos(4x) -(1/16)e ^{-x}Sin(4x)]\]

tada!

Answer 5

lol guessed you figured it out before i finished typing. Good job!