Question

Evaluate : $\int\limits^{\pi/2}_{\pi/4} {\sqrt{1-\sin 2x}} \, dx$

Answer 1

Solution:

As we know that
${sin}^{2} \: x + {cos}^{2} \: x = 1 \\ \\ sin \: 2x = 2sin \: x \: cos \: x$
put these values in the given function and convert it into integrable form

${\sqrt{1-\sin 2x}} = \sqrt{ \:{sin}^{2} \: x + {cos}^{2} \: x - 2 \: sin \: x \: cos \: x}$
now this can be written as

$\sqrt{ \:{sin}^{2} \: x + {cos}^{2} \: x - 2 \: sin \: x \: cos \: x} \\ \\ = \sqrt{ {(sin \: x - cos \: x})^{2} } \\ \\ = sin \: x - \: cos \: x$

$\int\limits^{\pi/2}_{\pi/4} {\sqrt{1-\sin 2x}} \, dx = \int\limits^{\pi/2}_{\pi/4} \: (sin \: x - \: cos \: x) \: dx \\ \\ = \int\limits^{\pi/2}_{\pi/4} \: sin \: x \: dx - \int\limits^{\pi/2}_{\pi/4} \:\: cos \: x \: dx \\ \\ = [- cos \: x \: - sin \: x]$

Now put the limits

$= - cos \: \frac{\pi}{2} +cos \: \frac{\pi}{4} \: - sin \: \frac{\pi}{2} + sin \: \frac{\pi}{4} \\ \\ = - 0 + \frac{1}{ \sqrt{2} } - 1 + \frac{1}{ \sqrt{2} } \\ \\ = \frac{2}{ \sqrt{2} } - 1 \\ \\ \int\limits^{\pi/2}_{\pi/4} {\sqrt{1-\sin 2x}} \, dx= \frac{2 - \sqrt{2} }{ \sqrt{2} }$