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If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x \,d x$ is equal to:
$2$
$-1$
$0$
$1$
Solution
$z=\frac{1}{1-\cos \theta+2 i \sin \theta}$
$=\frac{2 \sin ^{2} \frac{\theta}{2}-2 i \sin \theta}{(1-\cos \theta)^{2}+4 \sin ^{2} \theta}$
$=\frac{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}\left(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}\right)}$
$\operatorname{Re}(z)=\frac{1}{2\left(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}\right)}$
$=\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}=\frac{5}{2}$
$=1-\cos ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}=\frac{5}{2}$
$=3 \cos ^{2} \frac{\theta}{2}=\frac{3}{2}$
$=\cos ^{2} \frac{\theta}{2}=\frac{1}{2}$
$\frac{\theta}{2}=\mathrm{n} \pi \pm \frac{\pi}{4}$
$\theta=2 n \pi \pm \frac{\pi}{2}$
$\theta \in(0, \pi)$
$\theta=\frac{\pi}{2}$
$\int_{0}^{\frac{\pi}{2}} \sin \theta d \theta-[-\cos \theta]_{0}^{\frac{\pi}{2}}$
$=-(0-1)=1$