How to solve ∫2e^-2/tsin(3t)dt using integration by parts?

[xzibition8612]

Homework Statement

see attachment.

Homework Equations

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.

 

[Ray Vickson]

Homework Statement

see attachment.

Homework Equations

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.

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'.

 

[xzibition8612]

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

 

[HallsofIvy]

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

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.

 

[SammyS]

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 .​