Naughty Radical

Relevant wiki: Integration of Trigonometric Functions - Intermediate

Let $\large I = \int_0^{\pi /2} \sqrt{1- \sin2x} \, dx$

Now, $\large\sin 2x = 2\sin x\cos x\\\large1 = \sin^2x+\cos^2x\\\large\Rightarrow 1-\sin 2x = \left(\sin x-\cos x\right)^2\\\large\Rightarrow \sqrt{1-\sin 2x} = \large\left|\sin x- \cos x\right|$

So, $\displaystyle\large \begin{aligned}I &= \int_0^{\pi/2} \left|\sin x - \cos x\right|\, dx\\&=\int_0^{\pi/4} \left( \cos x- \sin x \right) \, dx + \int_{\pi/4}^{\pi/2} \left( \sin x - \cos x \right) \, dx\\ &=\left[\sin x + \cos x\right]_0^{ \pi/4} +\left[\cos x+\sin x \right]_{\pi/2}^{\pi/4}\\&= \dfrac1{\sqrt 2}+\dfrac1{\sqrt 2}-1+\dfrac1{\sqrt 2}+\dfrac1{\sqrt 2}-1 \\&= \boxed{2\sqrt2-2}\end{aligned}$

First of all, by symmetry of the sine function, $\mathfrak I = \int_0^{\pi/2} \sqrt{1 - \sin 2x}\:dx = 2\int_0^{\pi/4} \sqrt{1 - \sin 2x}\:dx.$

Switch variables: $u = \sin 2x$ so that $x = \tfrac12\arcsin u$ and $dx = 1/2\sqrt{1-u^2}\:du$ : $\mathfrak I = 2\int_0^1 \sqrt{1-u}\:\frac{du}{2\sqrt{1-u^2}} = \int_0^1 \frac{\sqrt{1-u}}{\sqrt{1-u^2}}\:du \\ = \int_0^1 \frac{du}{\sqrt{1+u}} = \left[2\sqrt{1+u}\right]_0^1 \\ = 2\sqrt 2 - 2.$ Thus the answer is $2 + 2 + 2 = \boxed{6}$ .

\begin{aligned} I & = \int_0^{\frac{\pi}{2}} \sqrt{1-\sin (2x)} \ dx \\ & = \int_0^{\frac{\pi}{2}} \sqrt{1-\cos \left(\color{#3D99F6}{\frac{\pi}{2} - 2x} \right)} \ dx \quad \quad \small \color{#3D99F6}{\text{Let }2u = \frac{\pi}{2} - 2x \implies du = - dx} \\ & = \int_{-\frac{\pi}{4}}^\frac{\pi}{4} \sqrt{1-\cos \left(\color{#3D99F6}{2u} \right)} \ du \quad \quad \small \color{#3D99F6}{\text{Since } \sqrt{1-\cos \left(2u \right)} \text{ is an even function}} \\ & = \color{#3D99F6}{2} \int_\color{#3D99F6}{0}^\frac{\pi}{4} \sqrt{1-\cos \left(2u \right)} \ du \\ & = 2 \int_0^\frac{\pi}{4} \sqrt{1-2\cos ^2 u +1} \ du \\ & = 2 \int_0^\frac{\pi}{4} \sqrt{2(1-\cos ^2 u)} \ du \\ & = 2\sqrt{2} \int_0^\frac{\pi}{4} \sin u \ du \\ & = 2\sqrt{2} (-\cos u) \ \bigg|_0^\frac{\pi}{4} \\ & = 2\sqrt{2} \left(-\frac{1}{\sqrt{2}} + 1\right) \\ & = 2\sqrt{2} - 2 \end{aligned}

$\implies a + b + c = 2+2+2 = \boxed{6}$

Another solution

$\begin{aligned} I & = \int_0^{\frac{\pi}{2}} \sqrt{1-\sin (2x)} \ dx \\ & = \int_0^{\frac{\pi}{2}} \sqrt{1-\frac{2 \tan x}{1+\tan^2 x}} \ dx \\ & = \int_0^{\frac{\pi}{2}} \sqrt{\frac{1+\tan^2 x - 2 \tan x}{1+\tan^2 x}} \ dx \\ & = \int_0^{\frac{\pi}{2}} \sqrt{\frac{(\tan x-1)^2}{\sec^2 x}} \ dx \\ & = \int_0^{\frac{\pi}{2}} \left| \frac{\tan x-1}{\sec x} \right| \ dx \\ & = \int_0^{\frac{\pi}{2}} \left| \sin x - \cos x \right| \ dx \\ & = \color{#3D99F6}{2} \int_0^{\color{#3D99F6}{\frac{\pi}{4}}} (\cos x-\sin x) \ dx \\ & = 2(\sin x + \cos x) \ \bigg|_0^\frac{\pi}{4} \\ & = 2\sqrt{2} - 2 \end{aligned}$

$\implies a + b + c = 2+2+2 = \boxed{6}$

$\displaystyle \int_{0}^{\pi/2} \sqrt{1 - \sin 2x} \, dx$

$\displaystyle = \int_{0}^{\pi/2} \dfrac{\sqrt{1- \sin 2x} \sqrt{1 + \sin 2x}}{\sqrt{1+\sin 2x}} \, dx$

$\displaystyle = \int_{0}^{\pi/2} \dfrac{\sqrt{1 - \sin^{2} 2x }}{\sqrt{1 + \sin 2x}}\, dx$

$\displaystyle = \int_{0}^{\pi/2} \dfrac{\left| \cos 2x \right| }{\sqrt{1+ \sin 2x }} \, dx$

$\displaystyle = 2 \int_{0}^{\pi/4} \dfrac{\cos 2x}{\sqrt{1 + \sin 2x}} \, dx$

Let $u = 1 + \sin 2x$ then $\dfrac{du}{dx} = 2 \cos 2x$

$\displaystyle = \int_{x = 0}^{x = \pi/4} \dfrac{du}{\sqrt{u}}$

$\displaystyle = \left. 2\sqrt{u} \right|_{x = 0}^{x= \pi/4}$

$\displaystyle = \left. 2 \sqrt{1 + \sin 2x} \right|_{0}^{\pi/4} = \boxed{2 \sqrt{2} - 2} \to 2+2+2 = \boxed{6}$