If the real part of the complex number $(1-\cos \mathrm{\theta }+2\mathrm{isin}\mathrm{\theta }{)}^{-1}\text{ is }\frac{1}{5}\text{ for }\mathrm{\theta }\in (0,\mathrm{\pi })$, then the value of the integral ${\int }_0^{\mathrm{\theta }} \sin \mathrm{xdx}$ is equal to

Let $\mathrm{z}=(1-\cos \mathrm{\theta }+2\mathrm{isin}\mathrm{\theta }{)}^{-1}$

$\begin{array}{c}\Rightarrow \mathrm{z}=\frac{1}{1-\cos \mathrm{\theta }+2\mathrm{isin}\mathrm{\theta }}\\ =\frac{1}{1-\cos \mathrm{\theta }+2\mathrm{isin}\mathrm{\theta }}\times \frac{1-\cos \mathrm{\theta }-2\mathrm{isin}\mathrm{\theta }}{1-\cos \mathrm{\theta }-2\mathrm{isin}\mathrm{\theta }}\\ =\frac{(1-\cos \mathrm{\theta })-2\mathrm{isin}\mathrm{\theta }}{(1-\cos \mathrm{\theta }{)}^2-(2\mathrm{isin}\mathrm{\theta }{)}^2}\\ =\frac{2{\sin }^2\frac{\mathrm{\theta }}{2}-4\mathrm{isin}\frac{\mathrm{\theta }}{2}\cos \frac{\mathrm{\theta }}{2}}{4{\sin }^4\frac{\mathrm{\theta }}{2}+16{\sin }^2\frac{\mathrm{\theta }}{2}{\cos }^2\frac{\mathrm{\theta }}{2}}\\ =\frac{2\sin \frac{\mathrm{\theta }}{2}\left(\sin \frac{\mathrm{\theta }}{2}-2\mathrm{icos}\frac{\mathrm{\theta }}{2}\right)}{4{\sin }^2\frac{\mathrm{\theta }}{2}\left({\sin }^2\frac{\mathrm{\theta }}{2}+4{\cos }^2\frac{\mathrm{\theta }}{2}\right)}\\ =\frac{\sin \frac{\mathrm{\theta }}{2}-2\mathrm{icos}\frac{\mathrm{\theta }}{2}}{2\sin \frac{\mathrm{\theta }}{2}\left({\sin }^2\frac{\mathrm{\theta }}{2}+4{\cos }^2\frac{\mathrm{\theta }}{2}\right)}\end{array}$

$\begin{array}{c}\mathrm{Now}, \mathrm{Re}\left(\mathrm{z}\right)=\frac{\sin \frac{\mathrm{\theta }}{2}}{2\sin \frac{\mathrm{\theta }}{2}\left({\sin }^2\frac{\mathrm{\theta }}{2}+4{\cos }^2\frac{\mathrm{\theta }}{2}\right)}\\ =\frac{1}{2\left(1+3{\cos }^2\frac{\mathrm{\theta }}{2}\right)}\end{array}$

Given, $\mathrm{Re}\left(\mathrm{z}\right)=\frac{1}{5}$

$\begin{array}{l}\Rightarrow \frac{1}{2\left(1+3{\cos }^2\frac{\mathrm{\theta }}{2}\right)}=\frac{1}{5}\\ \Rightarrow 1+3{\cos }^2\frac{\mathrm{\theta }}{2}=\frac{5}{2}\Rightarrow {\cos }^2\frac{\mathrm{\theta }}{2}=\frac{1}{2}\\ \Rightarrow \cos \frac{\mathrm{\theta }}{2}=\pm \frac{1}{\sqrt{2}}\\ \therefore \frac{\mathrm{\theta }}{2}=\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{4}\\ \therefore \mathrm{\theta }=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{2}\end{array}$

Given, range is $\mathrm{\theta }\in (0,\mathrm{\pi })$

$\therefore \mathrm{\theta }=\frac{\mathrm{\pi }}{2}$

$\mathrm{Now}, {\int }_0^{\mathrm{\theta }} \sin \mathrm{xdx}={\int }_0^{\frac{\mathrm{\pi }}{2}} \sin \mathrm{xdx}$

$\begin{array}{c}{\int }_0^{\frac{\mathrm{\pi }}{2}} \sin \mathrm{xdx}=-\cos \mathrm{x}{]}_0^{\mathrm{\pi }/2}=-\left(\cos \frac{\mathrm{\pi }}{2}-\cos 0\right)\\ =-(0-1)=1\end{array}$