Question

Evaluate: $\int\limits_{0}^{\frac{\pi}{4 }} \sqrt{1-sin\, 2x} \,\,\, dx$

• A. $\sqrt{2}-1$
• B. $\sqrt{2}+1$
• C. $\sqrt{2}$
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}-1$

Answer 2

We have, $I= \int\limits_{0}^{\frac{\pi}{4}}\sqrt{1-sin\, 2x } \,dx$ $= \int\limits_{0}^{\frac{\pi }{4}}\sqrt{sin ^{2}\,x +cos^{2}\,x -2\,sin \,x \,cos\, x }\,dx$ $=\int\limits_{0}^{\frac{\pi }{4}}\sqrt{\left(cos\,x -sin\,x\right)^{2}}= \int_{0}^{\frac{\pi }{4}}\left|\left(cos\,x-sin\, x\right)\right| dx$ $= \int\limits_{0}^{\pi/4}\left(cos\, x-sin\,x\right)dx \left[\because 0 < x < \pi/4, cos\,x > sin\,x \right]$ $= \left[sin\,x +cos \,x\right]_{0}^{\pi/4} = \frac{2}{\sqrt{2}} -1= \sqrt{2}-1$