Mathematics

Find $$\int { \dfrac { 1 }{ 1+tanx } dx }$$.


SOLUTION
$$\begin{array}{l} \displaystyle \int{ \dfrac { { dx } }{ { 1+\tan  x } }  }  \\ =\displaystyle \int{ \dfrac { { \cos{x}dx } }{ { \cos  x+\sin  x } }  }  \\ =\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { \left( { \sin  x+\cos  x } \right) +\left( { \cos  x-\sin  x } \right)  } }{ { \left( { \sin  x+\cos  x } \right)  } }  } dx \\ =\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { \left( { \cos  x-\sin  x } \right)  } }{ { \sin  x+\cos  x } }  } dx \\ \sin  x+\cos  x=t \\ =\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \displaystyle \int { \dfrac { { dt } }{ t }  }  \\ I=\dfrac { x }{ 2 } +\dfrac { 1 }{ 2 } \log  \left( { \sin  x+\cos  x } \right) +C \end{array}$$

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Subjective Medium Published on 17th 09, 2020
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