If the real part of the complex number ${(1 - \cos \theta + 2i\sin \theta )^{ - 1}}$ is ${1 \over 5}$ for $\theta \in (0,\pi )$, then the value of the integral $\int_0^\theta {\sin x} dx$ is equal to:

$z = {1 \over {1 - \cos \theta + 2i\sin \theta }}$<br><br>$$ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}$$<br><br>$$ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}$$<br><br>$${\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}$$<br><br>$\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}$$1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}$<br><br>$3{\cos ^2}{\theta \over 2} = {1 \over 2}$<br><br>${\theta \over 2} = n\pi \pm {\pi \over 4}$<br><br>$\theta = 2n\pi \pm {\pi \over 2}$<br><br>$\theta \in (0,\pi )$<br><br>$\therefore$ $\theta = {\pi \over 2}$<br><br>$$\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} $$<br><br>$= - (0 - 1)$<br><br>$= 1$