Mathematics

The value of $$\displaystyle \int_{0}^{\pi /4}\sqrt{\tan \theta }\: d\theta $$ is?


ANSWER

None of the above 


SOLUTION
Let $$\displaystyle I=\int { \sqrt { \tan { \theta  }  } d\theta  } =\frac { 1 }{ 2 } \left( { I }_{ 1 }-{ I }_{ 2 } \right) $$

Where 
$$\displaystyle { I }_{ 1 }=\int { \left( \tan { \theta  } +\cot { \theta  }  \right) d\theta  } =\int { \frac { \sin { \theta  } +\cos { \theta  }  }{ \sqrt { \cos { \theta  } \sin { \theta  }  }  } d\theta  } $$
Put $$\sin { \theta  } +\cos { \theta  } =u$$
$$\displaystyle { I }_{ 1 }=\sqrt { 2 } \int { \frac { dt }{ \sqrt { { 1-t }^{ 2 } }  }  } =\sqrt { 2 } \sin ^{ -1 }{ t } \\ =\sqrt { 2 } \sin ^{ -1 }{ \left( \sin { \theta  } -\cos { \theta  }  \right)  } $$

And  $${ I }_{ 2 }=\int  \left( \sqrt { \cot  \theta  } -\sqrt { \tan  \theta  }  \right) d\theta $$

$$\displaystyle =\int { \frac { \cos { \theta  } -\sin { \theta  }  }{ \sqrt { \cos { \theta  } \sin { \theta  }  }  } d\theta  } $$

Put $$\sin { \theta  } +\cos { \theta  } =t\Rightarrow 2\sin { \theta  } \cos { \theta  } ={ t }^{ 2 }-1$$
$$\displaystyle { I }_{ 2 }=\sqrt { 2 } \int { \frac { dt }{ \sqrt { { t }^{ 2 }-1 }  }  } =\sqrt { 2 } \log { \left( 1+\sqrt { { t }^{ 2 }-1 }  \right)  } \\ =\sqrt { 2 } \log { \left( \sin { \theta  } +\cos { \theta  } +\sqrt { \sin { 2\theta  }  }  \right)  } \\ I=\sqrt { 2 } \sin ^{ -1 }{ \left( \sin { \theta  } -\cos { \theta  }  \right)  } +\sqrt { 2 } \log { \left( \sin { \theta  } +\cos { \theta  } +\sqrt { \sin { 2\theta  }  }  \right)  } $$

Therefore
$$\displaystyle \int _{ 0 }^{ \pi /4 }{ Id\theta  } =\sqrt { 2 } \log { \left( \sqrt { 2 } +1 \right)  } -\frac { 3\pi  }{ 2 } $$
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Single Correct Hard Published on 17th 09, 2020
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