Can anyone help to integrat this function please?
$ ic \int_{-1}^{1} \left(\frac{-2}{x} \frac{1} {1+(\frac{tc}{x})^2}\right)dt $
C>0, x>0 and i the imaginary part
Not really sure how to go about integrating it.
Thanks
Can anyone help to integrat this function please?
$ ic \int_{-1}^{1} \left(\frac{-2}{x} \frac{1} {1+(\frac{tc}{x})^2}\right)dt $
C>0, x>0 and i the imaginary part
Not really sure how to go about integrating it.
Thanks
If $x$ is independent of t, and we have no reason to suppose it isn't, then we can factor it out of the integral and make the substitution $u=\frac{tc}{x}$, as suggested by Gregor Bruns, to get: $ic\int_{-1}^{1}\frac{-2}{x} \frac{1}{1+\left(\frac{tc}{x}\right)^{2}}=\frac{-2ic}{x}\int_{-c/x}^{c/x}\frac{1}{1+u^{2}}\cdot\frac{x}{c}du=-2i\int_{-c/x}^{c/x}\frac{1}{1+u^{2}}du$ Now you should recognise the integrand as $\frac{d}{du}\tan^{-1}(u)$, so we have $-2i\left[\tan^{-1}(c/x)-\tan^{-1}(-c/x)\right]=-4i\tan^{-1}(c/x)$