Repeating integration by parts

cameuth
Messages
17
Reaction score
0

Homework Statement



integrate .5e^(t/50)*sin(t)

Homework Equations


integration by parts
uv-∫vdu

The Attempt at a Solution



I am currently in differential equations and I remember from cal II that I have to keep using the equation above until the integral loops around, then set it equal to something,... but I'm foggy on what it's equal to. Any help would be appreciated. thanks.
 
Last edited by a moderator:
Physics news on Phys.org
Let's call the integral I. When you do integration by parts repeatedly, you eventually end up with

I = stuff + (something) I

At this point, you solve for I.
 
I believe that you should keep integrating by parts until you end up with something that looks like:

∫.5e^(t/50)*sin(t) = SOMETHING - ∫.5e^(t/50)*sin(t)

then you can add ∫.5e^(t/50)*sin(t) to both sides to get:

2*∫.5e^(t/50)*sin(t) = SOMETHING

so that

∫.5e^(t/50)*sin(t) = (1/2)*SOMETHING

EDIT: vela beat me to it :smile:
 
thanks. vela helped, and saladsamurai's elaboration was necessary. You guys are lifesavers.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top