$\displaystyle \int_{0}^{\pi /2} \sqrt{(1- \sin 2x)}=\\ a)2\sqrt{2} \; \; \; b)2(\sqrt{2}+1)\; \; \; c) 2 \; \; \; d)2(\sqrt{2}-1)$

 

Method 1

In order to find solution consider the properties of definite integral

Property1: 

$\displaystyle \int_{0}^{2a}f(x) \; dx =\begin{cases} 2\int_{0}^{a}f(x)\; dx & \text{ if } f(2a-x)=f(x) \\ 0 & \text{ if } f(2a-x)=-f(x) \end{cases}$ 

Property 2: 

$\displaystyle \int_{0}^{a} f(x) \; dx =\int_{0}^{a} f(a-x)\; dx$ 

Now consider the property 1, take $f(x)=\sqrt{1-\sin 2x}$ , $\displaystyle a= \frac{\pi }{4}$  and

$f(2a-x)=f(2\frac{\pi }{4} -x)=f(\frac{\pi }{2}-x)$ .

Here 

\[\begin{align*} f(2a-x) &=f(\frac{\pi }{2}-x ) \\ &= \sqrt{1-\sin 2(\frac{\pi }{2}-x)}\\ &= \sqrt{1-\sin (\pi -2x)}\\ &=\sqrt{1-\sin 2x} \; \; (\sin (\pi -x)=\sin x)\\ &= f(x) \end{align*}\]

Hence $\displaystyle \int_{0}^{2a}f(x)\; dx=2\int_{0}^{a}f(x)dx$ and

 \[\begin{align*} \int_{0}^{2\frac{\pi}{4}=\frac{\pi }{2}}\sqrt{1-\sin 2x}\; dx &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin 2x}\; dx \\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin 2(\frac{\pi }{4}-x)}\; dx \; (by\; property \; 2 )\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\sin (\frac{\pi }{2}-2x)}\; dx \\&=2\int_{0}^{ \frac{\pi}{4}}\sqrt{1-\cos 2x}\; dx \: \; (\because \sin (\frac{\pi }{2}-2x)= \cos 2x)\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{2\sin ^{2} x}\; dx \; \; (\because 1-\cos 2x= 2\sin ^{2} x )\\ &=2\int_{0}^{ \frac{\pi}{4}}\sqrt{2}\sin x \; dx \; \; (\because \sqrt{\sin ^{2}x}=\sin x\; for\; 0\leq x\leq \frac{\pi }{4})\\ &=2\sqrt{2}\left [ -\cos x \right ]_{0}^{\frac{\pi }{4}} \\ &=2\sqrt{2} \left [ -\cos \frac{\pi }{4}+\cos 0 \right ]\\ &= 2\sqrt{2}[-\frac{1}{\sqrt{2}}+1]\\&= 2[-1+\sqrt{2}]\\&= 2[\sqrt{2}-1] \end{align*}\]

Hence correct answer:(d)

Method 2

In order to find solution consider the definite integral

\[\begin{align*} \int_{0}^{\frac{\pi }{2}} \sqrt{(1-\sin 2x)}\: dx &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin ^{2} x+\cos ^{2} x-\sin 2x)} \; dx \; \; (\because 1=\sin ^{2} x+\cos ^{2} x)\\ &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin ^{2} x+\cos ^{2} x-2\sin x\cos x)}\; dx \; \; (\sin 2x=2\sin x\cos x)\\ &=\int_{0}^{\frac{\pi }{2}} \sqrt{(\sin x-\cos x)^{2}}\; dx \\ &= \int_{0}^{\frac{\pi }{2}} \left |(\sin x-\cos x) \right |\; dx \end{align*}\]

Here we have value of x ranges from  $0\leq x\leq \frac{\pi }{2}$ .

So we can examine that :

When $\displaystyle 0\leq x\leq \frac{\pi }{4} \; \; [\sin x -\cos x] \; \mathbf{is \; negative }.$ and

When $\displaystyle \frac{\pi }{4}\leq x\leq \frac{\pi }{2} \; \; [\sin x -\cos x] \; \mathbf{is \; positive }.$ 

Hence

\[\begin{align*} \int_{0}^{\frac{\pi }{2}}\left | \sin x - \cos x \right | dx&=\int_{0}^{\frac{\pi }{4}}-(\sin x - \cos x ) dx+\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\sin x - \cos x ) dx \\ &=-\left [ (-\cos x -\sin x) \right ]_{0}^{\frac{\pi }{4}}+[-\cos x - \sin x ]_{\frac{\pi }{4}}^{\frac{\pi }{2}} \\ &=-[(-\cos \frac{\pi }{4}-\sin \frac{\pi }{4})-(-\cos 0-\sin 0)]+[(-\cos \frac{\pi }{2}-\sin \frac{\pi }{2})-(-\cos \frac{\pi }{4}-\sin \frac{\pi }{4})] \\ &=- [(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})-(-1-0)]+[(-0-1)-(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})]\\ &=-[-\frac{2}{\sqrt{2}}+1]+[-1+\frac{2}{\sqrt{2}}]\\ &=-2+2\frac{2}{\sqrt{2}} \end{align*}\]

\[\therefore \int_{0}^{\frac{\pi }{2}}\sqrt{(1-\sin 2x)}=\int_{0}^{\frac{\pi }{2}}\left | \sin x - \cos x \right | dx =-2+2\frac{2}{\sqrt{2}}=-2+2\sqrt{2} \\ (\because 2=(\sqrt{2})^{2})\]

Hence 

\[\int_{0}^{\frac{\pi }{2}}\sqrt{(1-\sin 2x)} =2\sqrt{2}-2 =2(\sqrt{2}-1)\]

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