0 / 2 1 - sin 2 x ? d x is equal to:

Question

$\int_{0}^{π / 2} \sqrt{1 - \sin 2 x} d x$ is equal to:

Smart Achievers · Accepted Answer

(4)

Ans. $2 (\sqrt{2} - 1)$

Explanation :

Let $I = \int_{0}^{π / 2} \sqrt{1 - \sin 2 x} d x$

$= \int_{0}^{π / 2} \sqrt{\cos^{2} x + \sin^{2} x - 2 \sin x \cos x} d x$

$= \int_{0}^{π / 2} \sqrt{{(\cos x - \sin x)}^{2}} d x$

$= \int_{0}^{π / 4} \sqrt{{(\cos x - \sin x)}^{2}} d x + \int_{π / 4}^{π / 2} \sqrt{{(\sin x - \cos x)}^{2}} d x$

$[∵ \cos x > \sin x for x \in (0, \frac{π}{4})]$

$[∵ \sin x > \cos x for x \in (\frac{π}{4}, \frac{π}{2})]$

$= {[\sin x + \cos x]}_{0}^{π / 4} + {[- \cos x - \sin x]}_{π / 4}^{π / 2}$

$= [\sin \frac{π}{4} + \cos \frac{π}{4} - \sin 0 - \cos 0] + [- \cos \frac{π}{2} - \sin \frac{π}{2} + \cos \frac{π}{4} + \sin \frac{π}{4}]$

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

$= 2 \sqrt{2} - 2 = 2 (\sqrt{2} - 1)$

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