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

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

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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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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