Integration by parts (2-x)cos(nPi/2)x?

Splint
Messages
10
Reaction score
0

Homework Statement


Hi,

I'm doing fouier transforms and I'm not sure how to integrate (2-x)cos(nPi/2)x, (1,2). Anyone able to help me out? Even the indefinite integral would be fine.

Homework Equations



The Attempt at a Solution


I guess u would be (2-x) and dv would be cos(nPi/2)x dx. I'm not sure how to handle (nPi/2) since it is not a constant. Does (nPi/2) stay with x at all times?

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
n pi/2 IS a constant during the integration with respect to x.
Yes, it will be fine if you integrate by parts. How is the function exactly defined? Are you sure that you wrote the cosine function correctly? .

ehild
 
Last edited:
Thanks Ehild,

It's actually a half range expansion. The function is:

f(x)= (1, 0<x<1
(2-x, 1=<x=<2

So I believe the cosine function is correct. I won't try and integrate it right now but I should be ok with it.

Many thanks
Splint
 
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...
Back
Top