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

Let

$z=\frac{1}{1-\cos \theta +2i\sin \theta }$

$\Rightarrow z=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{\left(1-\cos \theta +2i\sin \theta \right)\left(1-\cos \theta -2i\sin \theta \right)}$

$\Rightarrow z=\frac{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)-2i\sin \theta }{(1-\cos \theta {)}^2+4{\sin }^2\theta }$

Put,$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}, 1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

$\Rightarrow z=\frac{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)-4i\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)}{4{\sin }^4\left({\displaystyle \frac{\theta }{2}}\right)+16{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right){\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}$

Hence,

$Re\left(z\right)=\frac{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)}{4{\sin }^4\left({\displaystyle \frac{\theta }{2}}\right)+16{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right){\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}$

$\Rightarrow \frac{1}{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)+8{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}=\frac{1}{5}$

$\Rightarrow \frac{1}{2{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)+8{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}=\frac{1}{5}$

$\Rightarrow {\sin }^2\left(\frac{\theta }{2}\right)+4{\cos }^2\left(\frac{\theta }{2}\right)=\frac{5}{2}$

$\Rightarrow 1-{\cos }^2\left(\frac{\theta }{2}\right)+4{\cos }^2\left(\frac{\theta }{2}\right)=\frac{5}{2}$

$\Rightarrow 3{\cos }^2\frac{\theta }{2}=\frac{3}{2}$

$\Rightarrow {\cos }^2\left(\frac{\theta }{2}\right)=\frac{1}{2}$

$\Rightarrow \frac{\theta }{2}=n\pi \pm \frac{\pi }{4}, n\in Z$

$\Rightarrow \theta =2n\pi \pm \frac{\pi }{2}$

For $\theta \in (0,\pi )$, $\theta =\frac{\pi }{2}$.

Hence,

$\therefore {\int }_0^{\theta }\sin \theta \mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}\sin \theta \mathrm{d}\theta ={\left[-\cos \theta \right]}_0^{\frac{\pi }{2}}$

$=-\left(0-1\right)$

$=1$