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Question

$$\displaystyle \int \cot^{-1} (\sec x-{\it \tan x} )dx =\ldots\ldots$$


A
πx4x24+c
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B
πx4+x24+c
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C
x24+c
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D
x22+c
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Solution

The correct option is A $$\displaystyle \frac{\pi x}{4}+\frac{x^{2}}{4}+c$$
$$\displaystyle I= \int \cot^{-1} (\sec x-{\tan x} )dx$$

$$\displaystyle  =\int \cot^{-1} \left(\dfrac {1-\sin x}{\cos x}\right)$$

$$\displaystyle =\int  \cot ^{ -1 } \left(\dfrac { \cos{ \dfrac { x }{ 2 }  }-\sin { \dfrac { x }{ 2 }  }  }{ \cos{ \dfrac { x }{ 2 }  }+\sin { \dfrac { x }{ 2 }  }  } \right)dx$$

$$\displaystyle =\int  \cot ^{ -1 } \left(\dfrac { 1-\tan { \dfrac { x }{ 2 }  }  }{ 1+\tan { \dfrac { x }{ 2 }  }  } \right)dx $$

$$\displaystyle =\int  \cot ^{ -1 } \left(\tan { \left(\dfrac { \pi  }{ 4 } -\dfrac { x }{ 2 } \right)}\right)dx$$

$$\displaystyle =\int  \cot^{ -1 } \left(\cot\left(\dfrac { \pi  }{ 2 } -\dfrac { \pi  }{ 4 } +\dfrac { x }{ 2 } \right)\right)dx$$

$$\Rightarrow \displaystyle I= \int \left(\dfrac { \pi  }{ 4 } +\dfrac { x }{ 2 }\right) dx$$

$$\Rightarrow \displaystyle I= \dfrac{\pi x}{4}+\dfrac{x^{2}}{4}+c$$

Mathematics

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