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

1. Sep 13, 2014

### aristotle_sind

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

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

$\int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = \int_{-R}^0 \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx$

$\Rightarrow \int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = - \int_{0}^{-R} \frac{\cos mx}{x^2 + 1}\,dx + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx$

Can't solve it any further :(

2. Sep 13, 2014

### LCKurtz

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. Sep 13, 2014

### aristotle_sind

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

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

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

Last edited: Sep 13, 2014
4. Sep 13, 2014

### LCKurtz

Show us what happens when you let $x=-u$ in your integral$$\int_{-R}^0 \frac{\cos (mx)}{x^2+1}~dx$$

5. Sep 13, 2014

### aristotle_sind

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

6. Sep 13, 2014

### LCKurtz

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. Sep 13, 2014

### aristotle_sind

Oh .. That clears everything up .. Thanks a lot mate...