# Integration tough cookie

1. Jan 16, 2013

### xzibition8612

1. The problem statement, all variables and given/known data
see attachment.

2. Relevant equations

3. The attempt at a solution
ok so the equation is in the attachment. My question is how did ∫2e....etc. in the first line get solved and get the answer in the second line. I tried integration by parts and set 2e^-2/t as the u and sin(3t)dt as dv and then did the uv-∫vdu, but then i get a ecos(3t) form for the ∫vdu and if i kept doing integration by parts i get never ending esin/cos. Anybody can help me or tell me how to solve this much appreciated.

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2. Jan 16, 2013

### Ray Vickson

You cannot set u = 2e^(-2/t); instead, set u = 2 e^(-t/2).

In problems like this, the integral I can be found using integration by parts twice: you will get an equation of the form I = f1(t) + f2(t)*I, so you can solve for I. I am surprised you have not seen this before; it is one of the standard 'tricks'.

3. Jan 17, 2013

### xzibition8612

well i apologize if my lack of intelligence offended you, but thanks anyway.

4. Jan 17, 2013

### HallsofIvy

Ray Vickson said nothing about your intelligence. He only said that he was surprised that you had not seen, before, a particular, but commonly used, method for this kind of problem. That has to do with experience, not intelligence.

Last edited by a moderator: Jan 17, 2013
5. Jan 17, 2013

### SammyS

Staff Emeritus

You may have a typo in your working of the problem.

The stated problem in the attachment has $\displaystyle \ \ e^{-t/2}\ .\$ I your Original Post, you are letting u=2e-2/t rather than letting u=2e-t/2 .

In either case, du/dt ≠ e .

$\displaystyle \frac{d}{dt} e^{-t/2}=\frac{-1}{2}e^{-t/2}\ .\$

Considering the sign and the coefficient of the cosine term, I'm convinced that you should go the other way with integration by parts,
Let u=sin(3t) and dv=2e-2/tdt .​

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