Question

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

Answer 1

${\int }_{\pi /4}^{\pi /2}\sqrt{1-sin2x}={\int }_{\pi /4}^{\pi /2}\sqrt{si{n}^{2}x+co{s}^{2x}}-2sinxcosx$

$={\int }_{\pi /4}^{\pi /2}|sinx-cosx|$

Now from $\frac{\pi }{4}<x<\frac{\pi }{2}$

$sinx>cosx$

$\therefore {\int }_{\pi /4}^{\pi /2}sinx-cosx={|}_{\pi /4}^{\pi /2}-cosx-{|}_{\pi /4}^{\pi /2}sinx$

$=-(0-\frac{1}{\sqrt{2}})-(1-\frac{1}{\sqrt{2}})$

$=\frac{1}{\sqrt{2}}-1+\frac{1}{\sqrt{2}}=\sqrt{2}-1$