question_answer

Evaluate \[\int_{\pi /4}^{\pi /2}{\sqrt{1-\sin 2x}\,dx}.\]

Answer:

Let \[l=\int_{\pi /4}^{\pi /2}{\sqrt{1-\sin 2x}}\,dx\] \[=\int_{\pi /4}^{\pi /2}{\sqrt{{{\cos }^{2}}x+{{\sin }^{2}}x-2\sin x\cos x}}\,dx\] \[=\int_{\pi /4}^{\pi /2}{\sqrt{{{(\cos x-\sin x)}^{2}}}}\,dx\] \[=\int_{\pi /4}^{\pi /2}{|\cos x-\sin x|}\,dx\]                 \[[\because \sqrt{{{x}^{2}}}=\,\,|x|]\] \[=\int_{\pi /4}^{\pi /2}{-\,(\cos x-\sin x)}\,dx\] \[=\int_{\pi /4}^{\pi /2}{(\sin \,x-\cos \,x)\,dx=[-\cos \,x-\sin \,x]}_{\pi /4}^{\pi /2}\] \[=\left\{ -\cos \left( \frac{\pi }{2} \right)-\sin \left( \frac{\pi }{2} \right) \right\}-\left\{ -\cos \left( \frac{\pi }{4} \right)-\sin \left( \frac{\pi }{4} \right) \right\}\] \[=(0-1)-\left( -\frac{2}{\sqrt{2}} \right)=\sqrt{2}-1\]