If the real part of (1-+2i)^-1 is 15 for (0,), then the value of _0^ x,dx is equal to

Question

If the real part of $(1-\cos	heta+2i\sin	heta)^{-1}$ is $\dfrac15$ for $	heta\in(0,\pi)$, then the value of $\displaystyle\int_0^	heta\sin x\,dx$ is equal to

Accepted Answer

Working through the half-angle substitution, $\operatorname{Re}(z)=\dfrac1{2\left(1+3\cos^2\frac\theta2\right)}=\dfrac15\Rightarrow\cos^2\dfrac\theta2=\dfrac12\Rightarrow\theta=\dfrac\pi2$ (in range $(0,\pi)$).

$$\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}=1.$$

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