∫limits_0^π/2√(1-sin x)dx is equal to

The correct answer is Option (1) → $2(\sqrt{2}-1)$

Evaluate the definite integral:

$\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin x} \, dx$

Use identity:

$1 - \sin x = \left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2$

So:

$\sqrt{1 - \sin x} = \left|\cos \frac{x}{2} - \sin \frac{x}{2}\right|$

For $x \in [0, \frac{\pi}{2}]$, $\cos \frac{x}{2} \ge \sin \frac{x}{2}$, so absolute value is not required. Thus:

$\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$

Therefore:

$\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin x} \, dx = \int_0^{\frac{\pi}{2}} \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) dx$

Use substitution: $u = \frac{x}{2} \Rightarrow dx = 2du$, and when $x = 0 \Rightarrow u = 0$, $x = \frac{\pi}{2} \Rightarrow u = \frac{\pi}{4}$

Then:

$\int_0^{\frac{\pi}{2}} \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) dx = 2 \int_0^{\frac{\pi}{4}} (\cos u - \sin u)\, du$

$= 2 \left[ \sin u + \cos u \right]_0^{\frac{\pi}{4}}$

$= 2 \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} - \sin 0 - \cos 0 \right)$

$= 2 \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 0 - 1 \right) = 2(\sqrt{2} - 1)$