[tex]\int\limits_{\pi/4}^{\pi/2}\sqrt{1-sin2x}dx[/tex]

[Гост]

[tex]\int\limits_{\pi/4}^{\pi/2}\sqrt{1-sin2x}dx[/tex]=?

[Knowledge Greedy]

[tex]\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sqrt{1-sin2x}dx=[/tex]

[tex]=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sqrt{(sinx-cosx)^2}dx=[/tex]

[tex]=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}(sinx-cosx)dx=\sqrt{2}-1[/tex],

защото в интервала [tex]\left (\frac{\pi}{4} ;\frac{\pi}{2} \right )[/tex]
[tex]\Rightarrow \,\ sinx-cosx>0[/tex]