Question

Evaluate: $\underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1-sin2x}dx$

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

Answer 1

Answer: (A)

We have,

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1-sin2x}dx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{si{n}^{2}x+co{s}^{2}x-2sinxcosx}dx$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{{\left(cosx-sinx\right)}^2}={\int }_0^{\frac{\pi }{4}}\left|\left(cosx-sinx\right)\right|dx$

$=\underset{0}{\overset{\pi /4}{\int }}\left(cosx-sinx\right)dx\left[\because 0<x<\pi /4,cosx>sinx\right]$

$={\left[sinx+cosx\right]}_0^{\pi /4}=\frac{2}{\sqrt{2}}-1=\sqrt{2}-1$