- #1
aristotle_sind
- 6
- 0
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 :(
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 :(