Mathematics

Solve $$\displaystyle \int _{\pi/6}^{\pi/3}\cos^{2}x dx$$


SOLUTION

$$\begin{array}{l} \int _{ \pi /6 }^{ \pi /3 }{ { { \cos   }^{ 2 } }xdx }  \\ =\dfrac { 1 }{ 2 } \int _{ \pi /6 }^{ \pi /3 }{ { { \cos   }^{ 2 } }xdx }  \\ =\dfrac { 1 }{ 2 } \int _{ \pi /6 }^{ \pi /3 }{ \left( { 1+\cos  2x } \right) dx }  \\ =\dfrac { 1 }{ 2 } { \left[ { x+\dfrac { { \sin  2x } }{ 2 }  } \right] _{ \pi /6 } }^{ \pi /3 } \\ =\dfrac { 1 }{ 2 } \left[ { \left( { \dfrac { \pi  }{ 3 } +\dfrac { 1 }{ 2 } \sin  \dfrac { { 2\pi  } }{ 3 }  } \right) -\left( { \dfrac { \pi  }{ 6 } +\dfrac { 1 }{ 2 } +\sin  \dfrac { \pi  }{ 3 }  } \right)  } \right]  \\ =\dfrac { 1 }{ 2 } \left[ { \left( { \dfrac { \pi  }{ 3 } +\dfrac { 1 }{ 2 } \times \sin  \dfrac { \pi  }{ 3 }  } \right) -\left( { \dfrac { \pi  }{ 2 } +\dfrac { 1 }{ 2 } \times \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right)  } \right]  \\ =\dfrac { 1 }{ 2 } \left[ { \left( { \dfrac { \pi  }{ 3 } +\dfrac { 1 }{ 2 } \times \dfrac { { \sqrt { 3 }  } }{ 3 }  } \right) -\left( { \dfrac { \pi  }{ 2 } +\dfrac { { \sqrt { 3 }  } }{ 4 }  } \right)  } \right]  \\ =\dfrac { 1 }{ 2 } \left[ { \dfrac { \pi  }{ 3 } +\dfrac { { \sqrt { 3 }  } }{ 4 } -\dfrac { \pi  }{ 2 } -\dfrac { { \sqrt { 3 }  } }{ 4 }  } \right]  \\ =\dfrac { 1 }{ 2 } \left[ { \dfrac { { -\pi  } }{ 6 }  } \right]  \\ =\dfrac { { -\pi  } }{ { 12 } }  \end{array}$$

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Subjective Medium Published on 17th 09, 2020
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