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Question

Solve:
$$\displaystyle \int_{0}^{1} \sqrt {\dfrac {1-x}{1+x}dx}$$


Solution

We have,
$$\displaystyle \int { \sqrt { \dfrac { 1-x }{ 1+x }  } dx } $$

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

$$=\displaystyle \int { \dfrac { 1-x }{ \sqrt { 1+{ x }^{ 2 } }  } dx } $$

$$=\displaystyle \int \left [ { \dfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  } -\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } }  }  } \right ] dx$$

$$=\sin ^{ -1 }{ x } +\sqrt { 1-{ x }^{ 2 } } +c$$

Hence, this is the answer.

Mathematics

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