# how do i find this integral?

• Jan 29th 2009, 07:18 AM
lord12
how do i find this integral?
integral(sinte^(6/489)(t))
• Jan 29th 2009, 07:30 AM
lord12
I know that I cannot use integration by parts since it always loops.
• Jan 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: $\displaystyle \int e^{\frac{6}{489}t}\sin t\,dt$??

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 $\displaystyle \int e^{\frac{6}{489}t}\sin t\,dt= \text{ stuff } + \text{ constants }\cdot \int e^{\frac{6}{489}t}\sin t\,dt$

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

Does this make sense?
• Jan 29th 2009, 08:50 AM
Soroban
Hello, lord12!

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

• Jan 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 $\displaystyle k = \tfrac{6}{489}$

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

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

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

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

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

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

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

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

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

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

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