$\int \frac{1}{1+x^2} = arctan(x) + c$

$\int \frac{1}{1+x^2} = \int \frac{1}{(1+xi)(1-xi)}$

$= \int \frac{\frac{1}{2}}{(1-xi)} + \frac{\frac{1}{2}}{(1+xi)}$

$=\frac{i}{2}ln(1-xi)-\frac{i}{2}ln(1+xi) = arctan(x) + c$

$=\frac{i}{2}ln(1-0i)-\frac{i}{2}ln(1+0i) = arctan(0) + c$

$0 + 0 = 0 + c$

$0 = c$

$\frac{i}{2}ln(1-xi)-\frac{i}{2}ln(1+xi) = arctan(x)$

$\frac{i}{2}ln(\frac{1-xi}{1+xi})= arctan(x)$

$ln(\frac{1-xi}{1+xi})=-2iarctan(x)$

$e^{-2iarctan(x)} = \frac{1-xi}{1+xi}$

substitute tangent of theta in place of x

$e^{-2i\theta} = \frac{1-tan(\theta)i}{1+tan(\theta)i}$

$e^{-2i\theta} = \frac{1-\frac{sin(\theta)}{cos(\theta)}i}{1+\frac{sin(\theta)}{cos(\theta)}i}$

$e^{-2i\theta} = \frac{cos(\theta)-isin(\theta)}{cos(\theta)+isin(\theta)}$

$e^{-2i\theta} = \frac{(cos(\theta)-isin(\theta))^2}{(cos(\theta)+isin(\theta))(cos(\theta)-isin(\theta))}$

$e^{-2i\theta} = \frac{cos^2(\theta)-sin^2(\theta)+2sin(\theta)cos(\theta)i}{cos^2(\theta)+sin^2(\theta)}$

$e^{-2i\theta} = \frac{cos(2\theta)+isin(2\theta)}{1}$

substitute minus theta over 2 in place of theta

$e^{-2i\theta} = \frac{cos(2\theta)+isin(2\theta)}{1}$

$e^{i\theta} = cos(\theta)+isin(\theta)$

Quod erat demonstrandum