Mathematics

If $$\displaystyle I=\int\frac{dx}{\sqrt{\left ( x-\alpha  \right )\left ( \beta -x \right )}},\left ( \beta < \alpha  \right )$$ then value of I is


ANSWER

$$\displaystyle \sin^{-1}\left ( \frac{2x-\alpha -\beta }{\beta -\alpha } \right )+C$$


SOLUTION
Let $$\displaystyle I= \int { \frac { dx }{ \sqrt { \left( x-\alpha  \right) \left( \beta -x \right)  }  }  } $$

Putting $$\displaystyle t=\frac { 1 }{ 2 } \left( x-\alpha +x-\beta  \right)   \Rightarrow dt=dx$$

$$ t=\displaystyle x-\frac { 1 }{ 2 } \left( \alpha +\beta  \right) $$

So $$\displaystyle \left( x-\alpha  \right) \left( \beta -x \right) =\left( t+\frac { 1 }{ 2 } \left( \alpha +\beta  \right) -\alpha  \right) \left( \beta -\frac { 1 }{ 2 } \left( \alpha +\beta  \right) -t \right) $$

$$\left( x-\alpha  \right) \left( \beta -x \right) \displaystyle =\frac { 1 }{ 4 } { \left( \beta -\alpha  \right)  }^{ 2 }-{ t }^{ 2 }$$

$$I=\displaystyle  \int { \frac { dt }{ \sqrt { \frac { 1 }{ 4 } { \left( \beta -\alpha  \right)  }^{ 2 }-{ t }^{ 2 } }  }  } $$

We know that, $$\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx =\sin^{-1}\left(\dfrac {x}{a}\right)+c$$


$$\displaystyle I=-\sin ^{ -1 }{ \left( \frac { 2t }{ \beta -\alpha  }  \right)  } +c=\sin ^{ -1 }{ \left( \frac { 2x-\alpha -\beta  }{ \beta -\alpha  }  \right) +c } $$
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