Mathematics

If $$\displaystyle\int {\dfrac{1}{{1 + \sin x}}dx = \tan \left( {\frac{x}{2} + a} \right) + b} $$, then 


ANSWER

$$a = - \dfrac{\pi }{4},b \in R$$


SOLUTION
$$\displaystyle \int \dfrac {dx}{1+\sin x}=\tan \left (\dfrac {x}{2}+a \right)+b$$
$$=\displaystyle \int \dfrac {dx}{1+\sin x}\dfrac {(1-\sin x)}{(1-\sin x)}=\displaystyle \int \dfrac {(1-\sin x)}{\cos^2 x}dx$$
$$=\displaystyle \int (\sec^2 x-\tan x \sec x)dx$$
$$=\tan x-\sec x-c=$$
$$=\dfrac { \sin { x } -1 }{ \cos { x }  } +c\dfrac { 2\cos { \dfrac { x }{ 2 } \sin { \dfrac { x }{ 2 } -1 }  }  }{ \cos ^{ 2 }{ \left( \dfrac { x }{ 2 }  \right)  } -\sin ^{ 2 }{ \left( \dfrac { x }{ 2 }  \right)  }  } $$
$$=\dfrac { 2\cos { \dfrac { x }{ 2 }  } \sin { \dfrac { x }{ 2 }  } -\left( \sin ^{ 2 }{ \dfrac { x }{ 2 }  } +\cos ^{ 2 }{ \dfrac { x }{ 2 }  }  \right)  }{ \left( \cos { \dfrac { x }{ 2 }  } -\sin { \dfrac { x }{ 2 }  }  \right) \left( \cos { \dfrac { x }{ 2 }  } +\sin { \dfrac { x }{ 2 }  }  \right)  } =-\dfrac { \left( \cos { \dfrac { x }{ 2 }  } -\sin { \dfrac { x }{ 2 }  }  \right)  }{ \left( \cos { \dfrac { x }{ 2 }  } +\sin { \dfrac { x }{ 2 }  }  \right)  } $$
$$=\dfrac { -\left( \dfrac { 1 }{ \sqrt { 2 }  } \cos { \dfrac { x }{ 2 }  } +\dfrac { 1 }{ \sqrt { 2 }  } \sin { \dfrac { x }{ 2 }  }  \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi  }{ 4 }  \right)  }  }{ -\left( \dfrac { 1 }{ \sqrt { 2 }  } \cos { \dfrac { x }{ 2 }  } -\dfrac { 1 }{ \sqrt { 2 }  } \sin { \dfrac { x }{ 2 }  }  \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi  }{ 4 }  \right)  }  } =\tan { \left( \dfrac { x }{ 2 } -\dfrac { \pi  }{ 4 }  \right)  } $$
$$a=\dfrac { \pi  }{ 4 } ,b\in R$$
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Single Correct Medium Published on 17th 09, 2020
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