Thank you for the replies...
Well .. my bad .. that certainly wasn't a good way to ask for help and my equation looked weird. I've bookmarked the guidelines thread .. will refer to it before making a new thread in future. Thank you for that...
And I did get a root that was 5/(2+i) but...
The equation is: (appears while solving a trigonometric integral using residue theorem)
2Z2+iZ2-6Z+2-i
=(2+i)Z2-6Z+(2-i)
The roots are:
Z1=(2-i) and Z2=(2-i)/5
I can't write the equation in factored form.
If I simply write it like this:
{z-(2-i)}{5z-(2-i)}
It doesn't give the same...
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...
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...