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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 :(