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

We need to find the real part of $$(1 - \cos\theta + 2i\sin\theta)^{-1}$$ and set it equal to $$\frac{1}{5}$$.

Let $$z = 1 - \cos\theta + 2i\sin\theta$$. Then $$z^{-1} = \frac{\overline{z}}{|z|^2}$$.

The real part of $$z^{-1}$$ is $$\frac{\text{Re}(z)}{|z|^2} = \frac{1 - \cos\theta}{(1-\cos\theta)^2 + 4\sin^2\theta}$$.

We expand the denominator: $$(1-\cos\theta)^2 + 4\sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + 4\sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + 4(1 - \cos^2\theta)$$ $$= 5 - 2\cos\theta - 3\cos^2\theta.$$

Setting the real part equal to $$\frac{1}{5}$$: $$\frac{1-\cos\theta}{5 - 2\cos\theta - 3\cos^2\theta} = \frac{1}{5}.$$

Let $$c = \cos\theta$$. Then: $$5(1 - c) = 5 - 2c - 3c^2$$ $$5 - 5c = 5 - 2c - 3c^2$$ $$-5c = -2c - 3c^2$$ $$0 = 3c - 3c^2 = 3c(1 - c).$$

So $$c = 0$$ or $$c = 1$$. Since $$\theta \in (0, \pi)$$, we have $$\cos\theta \neq 1$$, so $$\cos\theta = 0$$, giving $$\theta = \frac{\pi}{2}$$.

Therefore, $$\int_0^\theta \sin x\, dx = \int_0^{\pi/2} \sin x\, dx = [-\cos x]_0^{\pi/2} = -\cos\frac{\pi}{2} + \cos 0 = 0 + 1 = 1.$$