how do i find this integral?

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January 29th 2009, 07:18 AM
lord12

how do i find this integral?

integral(sinte^(6/489)(t))
January 29th 2009, 07:30 AM
lord12

I know that I cannot use integration by parts since it always loops.
January 29th 2009, 07:46 AM
Chris L T521

Quote:

Originally Posted by lord12

I know that I cannot use integration by parts since it always loops.

Is this your integral: $\int e^{\frac{6}{489}t}\sin t\,dt$ ??

If so you can use integration by parts. The key to this one is that you apply integration by parts twice. After you apply it twice, you'll see the original integral reappear.

I.e. you'll see after applying IBP twice that you will end up with $\int e^{\frac{6}{489}t}\sin t\,dt= \text{ stuff } + \text{ constants }\cdot \int e^{\frac{6}{489}t}\sin t\,dt$

Then you treat $\int e^{\frac{6}{489}t}\sin t\,dt$ as a variable and then solve for it.

Does this make sense?
January 29th 2009, 08:50 AM
Soroban

Hello, lord12!

Yes, it "loops" when doing it by-parts, but that's what we want!

Quote:

$\int e^{\frac{6}{489}t}\sin t\,dt$ . Why did they use such an ugly fraction?

Let $k = \tfrac{6}{489}$

We have: . $I \;=\;\int e^{kt}\sin t\,dt$

. . By parts: . $\begin{array}{ccccccc}u &=& \sin t && dv &=& e^{kt}dt \\ du &=& \cos t\,dt & & v &=&\frac{1}{k}e^{kt} \end{array}$

And we have: . $I \;=\;\frac{1}{k}\,e^{kt}\sin t - \frac{1}{k}\int e^{kt}\cos t\,dt$

. . By parts: . $\begin{array}{ccccccc}u &=& \cos t && dv &=&e^{kt}dt \\ du&=& -\sin t\,dt && v &=&\frac{1}{k}e^{kt} \end{array}$

And we have: . $I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k}\left[\frac{1}{k}e^{kt}\cos t + \frac{1}{k}\int e^{kt}\sin t\,dt\right]$

. . . . . . . . . . $I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t - \frac{1}{k^2}\underbrace{\int e^{kt}\sin t\,dt}_{\text{This is }I} + C$

We have: . $I + \frac{1}{k^2}I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t + C$

Factor: . $\frac{k^2+1}{k^2}\,I \;=\;\frac{e^{kt}}{k^2}(k\sin t -\cos t) + C$

Then: . $I \;=\;\frac{k^2}{k^2+1}\cdot\frac{e^{kt}}{k^2}(k\si n t - \cos t) + C$

Therefore: . $I \;=\;\frac{e^{kt}}{k^2+1}(k\sin t - \cos t) + C$
January 29th 2009, 09:12 AM
heathrowjohnny

Quote:

Originally Posted by Soroban

Hello, lord12!

Yes, it "loops" when doing it by-parts, but that's what we want!

Let $k = \tfrac{6}{489}$

We have: . $I \;=\;\int e^{kt}\sin t\,dt$

. . By parts: . $\begin{array}{ccccccc}u &=& \sin t && dv &=& e^{kt}dt \\ du &=& \cos t\,dt & & v &=&\frac{1}{k}e^{kt} \end{array}$

And we have: . $I \;=\;\frac{1}{k}\,e^{kt}\sin t - \frac{1}{k}\int e^{kt}\cos t\,dt$

. . By parts: . $\begin{array}{ccccccc}u &=& \cos t && dv &=&e^{kt}dt \\ du&=& -\sin t\,dt && v &=&\frac{1}{k}e^{kt} \end{array}$

And we have: . $I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k}\left[\frac{1}{k}e^{kt}\cos t + \frac{1}{k}\int e^{kt}\sin t\,dt\right]$

. . . . . . . . . . $I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t - \frac{1}{k^2}\underbrace{\int e^{kt}\sin t\,dt}_{\text{This is }I} + C$

We have: . $I + \frac{1}{k^2}I \;=\;\frac{1}{k}e^{kt}\sin t - \frac{1}{k^2}e^{kt}\cos t + C$

Factor: . $\frac{k^2+1}{k^2}\,I \;=\;\frac{e^{kt}}{k^2}(k\sin t -\cos t) + C$

Then: . $I \;=\;\frac{k^2}{k^2+1}\cdot\frac{e^{kt}}{k^2}(k\si n t - \cos t) + C$

Therefore: . $I \;=\;\frac{e^{kt}}{k^2+1}(k\sin t - \cos t) + C$

because it was probably needed to solve a differential equation problem.