Mathematics

$$\int {(\tan x-\cot x)^2dx}=$$


ANSWER

$$\tan x-\cot x-4x+C$$


SOLUTION
$$I=\int { { \left( \tan { x } -\cot { x }  \right)  }^{ 2 } } dx=\int { \cfrac { { \left( \tan ^{ 2 }{ x } -1 \right)  }^{ 2 } }{ \tan ^{ 2 }{ x }  }  } dx=\int { \cfrac { \tan ^{ 4 }{ x } +1-2\tan ^{ 2 }{ x }  }{ \tan ^{ 2 }{ x }  }  } dx$$
$$=\int { \tan ^{ 2 }{ x }  } dx+\int { \cot ^{ 2 }{ x }  } dx-2\int { dx } =\int { \left[ \left( \tan ^{ 2 }{ x } +1 \right) -1 \right]  } dx+\int { \cfrac { \cos ^{ 2 }{ x }  }{ \sin ^{ 2 }{ x }  }  } dx-2x+C$$
$$=\int { \sec ^{ 2 }{ x }  } dx-\int { dx } +\int { \cfrac { 1-\sin ^{ 2 }{ x }  }{ \sin ^{ 2 }{ x }  }  } dx-2x+C$$
$$=\tan { x } -x+\int { co\sec ^{ 2 }{ x }  } dx-\int { dx } -2x+C=\tan { x } -4x+\int { co\sec ^{ 2 }{ x }  } dx+C=\tan { x } -4x-\cot { x } +C\quad $$
$$I=\tan { x } -\cot { x } -4x+C$$
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Single Correct Medium Published on 17th 09, 2020
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