$<br /> \int \frac{1}{1+x^2} = arctan(x) + c<br /> \newline<br /> \newline<br /> \int \frac{1}{1+x^2} = \int \frac{1}{(1+xi)(1-xi)}<br /> \newline<br /> \newline<br /> = \int \frac{\frac{1}{2}}{(1-xi)} + \frac{\frac{1}{2}}{(1+xi)}<br /> \newline<br /> \newline<br /> =\frac{i}{2}ln|1-xi|-\frac{i}{2}ln|1+xi| = arctan(x) + c<br /> \newline<br /> \newline<br /> =\frac{i}{2}ln|1-0i|-\frac{i}{2}ln|1+0i| = arctan(0) + c<br /> \newline<br /> \newline<br /> 0 + 0 = 0 + c<br /> \newline<br /> \newline<br /> 0 = c<br /> \newline<br /> \newline<br /> \frac{i}{2}ln|1-xi|-\frac{i}{2}ln|1+xi| = arctan(x)<br /> \newline<br /> \newline<br /> \frac{i}{2}ln|\frac{1-xi}{1+xi}|= arctan(x)<br /> \newline<br /> \newline<br /> ln|\frac{1-xi}{1+xi}|=-2iarctan(x)<br /> \newline<br /> \newline<br /> e^{-2iarctan(x)} = \frac{1-xi}{1+xi}<br /> \newline<br /> \newline<br />$
substitute tangent of theta in place of x
$<br /> e^{-2i\theta} = \frac{1-tan(\theta)i}{1+tan(\theta)i}<br /> \newline<br /> \newline<br /> e^{-2i\theta} = \frac{1-\frac{sin(\theta)}{cos(\theta)}i}{1+\frac{sin(\theta)}{cos(\theta)}i}<br /> \newline<br /> \newline<br /> e^{-2i\theta} = \frac{cos(\theta)-isin(\theta)}{cos(\theta)+isin(\theta)}<br /> \newline<br /> \newline<br /> e^{-2i\theta} = \frac{(cos(\theta)-isin(\theta))^2}{(cos(\theta)+isin(\theta))(cos(\theta)-isin(\theta))}<br /> \newline<br /> \newline<br /> e^{-2i\theta} = \frac{cos^2(\theta)-sin^2(\theta)+2sin(\theta)cos(\theta)i}{cos^2(\theta)+sin^2(\theta)}<br /> \newline<br /> \newline<br /> e^{-2i\theta} = \frac{cos(2\theta)+isin(2\theta)}{1}<br />$
substitute minus theta over 2 in place of theta
$<br /> e^{-2i\theta} = \frac{cos(2\theta)+isin(2\theta)}{1}<br /> \newline<br /> \newline<br /> e^{i\theta} = cos(\theta)+isin(\theta)<br />$
Quod erat demonstrandum