Mathematics

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


ANSWER

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


SOLUTION
Let $$I = \displaystyle \int_{0}^{\pi/2} \sqrt{1 - sin \,2x \,dx}$$
$$= \displaystyle \int_{0}^{\pi/4} \sqrt{(cos \,x - sin \,x)^2} dx + \displaystyle \int_{\pi/4}^{\pi/2} \sqrt{(sin x - cos x)^2} dx$$
$$= [sin x + cos x ]_{0}^{\pi/4} + [-cos x - sin x]_{\pi/4}^{\pi/2}$$
$$= \dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2} - 0 - 1 + \left ( -0 - 1 + \dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2} \right )$$
$$= 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$$
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Single Correct Medium Published on 17th 09, 2020
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