# Math Help - Integration by Parts - Matching with Original Integral

1. ## Integration by Parts - Matching with Original Integral

$\int e^{4x} sin(5x) dx$

I know I have to use integration by parts and match up with the original integral since neither function will get down to a coefficient.

$\int u dv = uv - \int v du$

Here $u = e^{4x}$ $dv = sin(5x)$ $du=4e^{4x}$ $v= - \frac{1}{5}cos(5x)$

$\int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) - \int -\frac{1}{5} cos(5x)(4e^{4x}) dx$

$\int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) + \frac{4}{5} \int e^{4x} cos(5x)dx$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[e^{4x}\left(\frac{1}{5}sin(5x)\right) - \int \frac{1}{5}sin(5x)(4e^{4x})dx\right]$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[\frac{1}{5}e^{4x}sin(5x) - \frac{4}{5} \int e^{4x}sin(5x)dx\right]$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) + \frac{4}{25}e^{4x}sin(5x) - \frac{16}{25} \int e^{4x}sin(5x)dx$

At this point I'm stuck. I can see that on both sides of the equations I do have the matching integral but the coefficient in front of the integral on the right side is tripping me up. Or perhaps I screwed up somewhere.

2. Originally Posted by VitaX
$\int e^{4x} sin(5x) dx$

I know I have to use integration by parts and match up with the original integral since neither function will get down to a coefficient.

$\int u dv = uv - \int v du$

Here $u = e^{4x}$ $dv = sin(5x)$ $du=4e^{4x}$ $v= - \frac{1}{5}cos(5x)$

$\int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) - \int -\frac{1}{5} cos(5x)(4e^{4x}) dx$

$\int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) + \frac{4}{5} \int e^{4x} cos(5x)dx$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[e^{4x}\left(\frac{1}{5}sin(5x)\right) - \int \frac{1}{5}sin(5x)(4e^{4x})dx\right]$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[\frac{1}{5}e^{4x}sin(5x) - \frac{4}{5} \int e^{4x}sin(5x)dx\right]$

$\int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) + \frac{4}{25}e^{4x}sin(5x) - \frac{16}{25} \int e^{4x}sin(5x)dx$

At this point I'm stuck. I can see that on both sides of the equations I do have the matching integral but the coefficient in front of the integral on the right side is tripping me up. Or perhaps I screwed up somewhere.
Move $- \frac{16}{25} \int e^{4x}sin(5x)dx$ to the left hand side, then take $\int e^{4x}sin(5x)dx$ as common factor..

3. Originally Posted by General
Move $- \frac{16}{25} \int e^{4x}sin(5x)dx$ to the left hand side, then take $\int e^{4x}sin(5x)dx$ as common factor..
ok I didn't think about common factors at the time. by the way isn't there a short cut to integrating by parts if you need to do it more than once?

4. Originally Posted by VitaX
by the way isn't there a short cut to integrating by parts if you need to do it more than once?
Yes. Reduction formula.