Mathematics

The value of $$\displaystyle \int_{0}^{3\pi /4}$$ $$\sqrt{1-sin2x} \ dx$$


ANSWER

$$2\sqrt{2}+1$$


SOLUTION
$$\displaystyle \int _{ 0 }^{ \frac { 3\pi  }{ 4 }  }{ \sqrt { 1-\sin { 2x }  }  } dx=\int _{ 0 }^{ \frac { 3\pi  }{ 4 }  }{ \sqrt { { \left( \sin { x } -\cos { x }  \right)  }^{ 2 } }  } dx$$
$$\displaystyle =\int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \left( \cos { x } -\sin { x }  \right) dx } +\int _{ \frac { \pi  }{ 4 }  }^{ \frac { 3\pi  }{ 4 }  }{ \left( \sin { x } -\cos { x }  \right)  } dx$$
$$\displaystyle =\left[ \sin { x } +\cos { x }  \right] _{ 0 }^{ \frac { \pi  }{ 4 }  }{ + }\left[ -\cos { x } -\sin { x }  \right] _{ \frac { \pi  }{ 4 }  }^{ \frac { 3\pi  }{ 4 }  }$$
$$\displaystyle =\frac { 1 }{ \sqrt { 2 }  } +\frac { 1 }{ \sqrt { 2 }  } +1+\frac { 1 }{ \sqrt { 2 }  } -\frac { 1 }{ \sqrt { 2 }  } +\frac { 1 }{ \sqrt { 2 }  } +\frac { 1 }{ \sqrt { 2 }  } \\ =2\sqrt { 2 } +1$$
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Single Correct Medium Published on 17th 09, 2020
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