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

1. Solve the trigonometric inequality for the given range.

$1 - 2\sin x > 0$
$\sin x < \frac{1}{2}$

For $0 \le x \le \frac{\pi}{2}$, we know $\sin \frac{\pi}{6} = \frac{1}{2}$.
Since $\sin x$ is an increasing function in the first quadrant, the solution is:
$0 \le x < \frac{\pi}{6}$

2. Split the definite integral using the sign of the expression inside the absolute value.

From the inequality solved, $1-2\sin x \ge 0$ for $0 \le x \le \frac{\pi}{6}$, and $1-2\sin x < 0$ for $\frac{\pi}{6} < x \le \frac{\pi}{2}$.

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

3. Integrate and evaluate the limits for both parts.

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

Evaluating the first part:
$\left(\frac{\pi}{6} + 2\cos\frac{\pi}{6}\right) - \left(0 + 2\cos 0\right) = \frac{\pi}{6} + \sqrt{3} - 2$

Evaluating the second part:
$\left(-2\cos\frac{\pi}{2} - \frac{\pi}{2}\right) - \left(-2\cos\frac{\pi}{6} - \frac{\pi}{6}\right)$
$= \left(0 - \frac{\pi}{2}\right) - \left(-\sqrt{3} - \frac{\pi}{6}\right) = \sqrt{3} - \frac{\pi}{3}$

4. Sum the components to find the final exact value.

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

1. Apply the given substitution and double-angle identity to transform the integral.

Let $x = \sin\theta$. Then differentiating gives $\frac{dx}{d\theta} = \cos\theta$, so $dx = \cos\theta \,d\theta$.
Using the double-angle identity, we have $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2x^2$.
Substituting these into the given integral:
$\int \frac{\cos\theta}{\sqrt{2\cos 2\theta -1}}\,d\theta = \int \frac{1}{\sqrt{2(1-2x^2)-1}}\,dx$
$= \int \frac{1}{\sqrt{2 - 4x^2 - 1}}\,dx$
$= \int \frac{1}{\sqrt{1 - 4x^2}}\,dx \quad \text{(shown)}$

2. Determine the new limits of integration and evaluate the definite integral in terms of $\alpha$.

For the definite integral, we change the limits of integration from $\theta$ to $x$:
When $\theta = 0$, $x = \sin 0 = 0$.
When $\theta = \alpha$, $x = \sin\alpha$.
Therefore, the integral becomes:
$\int_{0}^{\sin\alpha} \frac{1}{\sqrt{1 - 4x^2}}\,dx$
Using the standard result $\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\left(\frac{x}{a}\right)$:
$\int_{0}^{\sin\alpha} \frac{1}{\sqrt{1 - (2x)^2}}\,dx = \left[ \frac{1}{2}\sin^{-1}(2x) \right]_{0}^{\sin\alpha}$
$= \frac{1}{2}\sin^{-1}(2\sin\alpha) - \frac{1}{2}\sin^{-1}(0)$
$= \frac{1}{2}\sin^{-1}(2\sin\alpha)$

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

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