Solve the inequality $1-2\sin x>0$ for $0\le x\le \frac{\pi}{2}$. Hence, evaluate the exact value of $\displaystyle \int_{0}^{\frac{\pi}{2}} \big|1-2\sin x\big|\,dx$. Use the substitution $x=\sin\theta$ to show that $\displaystyle \int \frac{\cos\theta}{\sqrt{2\cos 2\theta -1}}\,d\theta = \int \frac{1}{\sqrt{1-4x^2}}\,dx$. Given that $\displaystyle \int_{0}^{\alpha}\frac{\cos\theta}{\sqrt{2\cos 2\theta -1}}\,d\theta = \frac{\pi}{4}$, find the value of $\alpha$, given that $0<\alpha<\frac{\pi}{2}$.

1. Solve the trigonometric inequality for the given domain.

$1 - 2\sin x > 0$
$\sin x < \frac{1}{2}$
For $0 \le x \le \frac{\pi}{2}$, the critical value is $x = \frac{\pi}{6}$.
Since $\sin x$ is increasing in this domain, the solution is:
$0 \le x < \frac{\pi}{6}$

2. Use the result of the inequality to split the definite integral and remove the absolute value.

Since $1-2\sin x > 0$ for $0 \le x < \frac{\pi}{6}$ and $1-2\sin x \le 0$ for $\frac{\pi}{6} \le x \le \frac{\pi}{2}$:

$\int_{0}^{\frac{\pi}{2}} |1-2\sin x|\,dx = \int_{0}^{\frac{\pi}{6}} (1-2\sin x)\,dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} -(1-2\sin x)\,dx$
$= \int_{0}^{\frac{\pi}{6}} (1-2\sin x)\,dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (2\sin x - 1)\,dx$

3. Perform the integration of the trigonometric functions.

$= \left[ x + 2\cos x \right]_0^{\frac{\pi}{6}} + \left[ -2\cos x - x \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

4. Evaluate the definite integral exactly by substituting the limits.

$= \left( \frac{\pi}{6} + 2\cos\frac{\pi}{6} \right) - (0 + 2\cos 0) + \left( -2\cos\frac{\pi}{2} - \frac{\pi}{2} \right) - \left( -2\cos\frac{\pi}{6} - \frac{\pi}{6} \right)$
$= \left( \frac{\pi}{6} + 2\left(\frac{\sqrt{3}}{2}\right) \right) - 2 + \left( 0 - \frac{\pi}{2} \right) - \left( -2\left(\frac{\sqrt{3}}{2}\right) - \frac{\pi}{6} \right)$
$= \left( \frac{\pi}{6} + \sqrt{3} - 2 \right) + \left( -\frac{\pi}{2} + \sqrt{3} + \frac{\pi}{6} \right)$
$= 2\sqrt{3} - 2 - \frac{\pi}{6}$

1. Differentiate the substitution and express the integrand's denominator in terms of $x$.

Given $x = \sin\theta$, differentiating gives:
$\frac{dx}{d\theta} = \cos\theta \implies dx = \cos\theta \, d\theta$

Using the double angle identity $\cos 2\theta = 1 - 2\sin^2\theta$, we substitute into the expression inside the square root:
$2\cos 2\theta - 1 = 2(1 - 2\sin^2\theta) - 1$
$2\cos 2\theta - 1 = 2 - 4\sin^2\theta - 1 = 1 - 4\sin^2\theta$
Since $x = \sin\theta$, this becomes $1 - 4x^2$.

2. Substitute into the integral to complete the "show that" part.

$\int \frac{\cos\theta}{\sqrt{2\cos 2\theta - 1}} \, d\theta = \int \frac{1}{\sqrt{1 - 4\sin^2\theta}} (\cos\theta \, d\theta)$
$= \int \frac{1}{\sqrt{1 - 4x^2}} \, dx$

3. Change the limits of integration and evaluate the definite integral using standard forms.

Change the limits of integration from $\theta$ to $x$:
When $\theta = 0$, $x = \sin(0) = 0$.
When $\theta = \alpha$, $x = \sin\alpha$.

Evaluating the definite integral:
$\int_0^{\alpha} \frac{\cos\theta}{\sqrt{2\cos 2\theta - 1}} \, d\theta = \int_0^{\sin\alpha} \frac{1}{\sqrt{1 - 4x^2}} \, dx$
$= \left[ \frac{1}{2} \sin^{-1}(2x) \right]_0^{\sin\alpha}$
$= \frac{1}{2} \sin^{-1}(2\sin\alpha) - \frac{1}{2}\sin^{-1}(0)$
$= \frac{1}{2} \sin^{-1}(2\sin\alpha)$

4. Equate the evaluated integral to $\frac{\pi}{4}$ and solve for $\alpha$.

We are given that the integral evaluates to $\frac{\pi}{4}$:
$\frac{1}{2} \sin^{-1}(2\sin\alpha) = \frac{\pi}{4}$
$\sin^{-1}(2\sin\alpha) = \frac{\pi}{2}$
$2\sin\alpha = \sin\left(\frac{\pi}{2}\right) = 1$
$\sin\alpha = \frac{1}{2}$
Given the restriction $0 < \alpha < \frac{\pi}{2}$:
$\alpha = \frac{\pi}{6}$