Changing limits of integration - definite integral (without u sub)?

  • #1
Hello
Can someone please tell me how is: [itex]\int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = 2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]
where,
m and R are positive real numbers

This is how I'm trying to solve it...

[itex]\int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = \int_{-R}^0 \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

[itex]\Rightarrow \int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = - \int_{0}^{-R} \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

Can't solve it any further :(
 

Answers and Replies

  • #2
LCKurtz
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It's true for any even function ##f(x)##. That means ##f(-x) = f(x)## so the graph is symmetric about the y axis. In your argument try ##x=-u## in your first integral at the end.
 
  • #3
It's true for any even function ##f(x)##. That means ##f(-x) = f(x)## so the graph is symmetric about the y axis. In your argument try ##x=-u## in your first integral at the end.

so I put ##x=-u## in one of the integrals .. this is what I get...

[itex]\int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = \int_{0}^{R} \frac{\cos mu}{u^2 + 1}\,du + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

but now the two integrals on RHS have different variables .. How do I add them so the resulting expression becomes
[itex]2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]
 
Last edited:
  • #4
LCKurtz
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Show us what happens when you let ##x=-u## in your integral$$
\int_{-R}^0 \frac{\cos (mx)}{x^2+1}~dx$$
 
  • #5
Show us what happens when you let ##x=-u## in your integral$$
\int_{-R}^0 \frac{\cos (mx)}{x^2+1}~dx$$

I just changed my previous post .. can you please check that...
 
  • #6
LCKurtz
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so I put ##x=-u## in one of the integrals .. this is what I get...

[itex]\int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = \int_{0}^{R} \frac{\cos mu}{u^2 + 1}\,du + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

but now the two integrals on RHS have different variables .. How do I add them so the resulting expression becomes
[itex]2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx[/itex]

I just changed my previous post .. can you please check that...

The ##x## and ##u## variables are dummy variables. You put the ##0## and ##R## limits in either and get the same answer from both.
 
  • #7
Oh .. That clears everything up .. Thanks a lot mate...
 

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