Mathematics

Evaluate: $$\displaystyle \int \dfrac{1}{1+tanx}dx$$


SOLUTION
$${l} \displaystyle \int{ \dfrac { { dx } }{ { 1+\tan  x } }  }  \\ $$


$$=\displaystyle \int{ \dfrac { { \cos{x}dx } }{ { \cos  x+\sin  x } }  }  \\$$

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

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

$$ \sin  x+\cos  x=t \\ $$

$$=\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { dt } }{ t }  }  \\ $$

$$I=\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \log  \left( { \sin  x+\cos  x } \right) +C $$
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