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Also sprach Zarathustra
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- Jan 31, 2012
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Calculate the following integral:
$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$
$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$
Calculate the following integral:
$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$
\[ \displaystyle \begin{align*} \int_a^b{\frac{dx}{\sqrt{(x - a)(b - x)}}} &= \int_a^b{\frac{dx}{\sqrt{-x^2 + (a + b)x - ab}}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left[ x^2 - (a + b)x + ab \right]}}} \\ &= \int{\frac{dx}{\sqrt{-\left\{ x^2 - (a + b)x + \left[-\left(\frac{a + b}{2}\right)\right]^2 - \left[-\left(\frac{a + b}{2}\right)\right]^2 + ab \right\} }}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left\{ \left[ x - \left( \frac{a+b}{2} \right) \right]^2 - \frac{(a + b)^2}{4} + \frac{4ab}{4} \right\}}}} \\ &= \int_a^b{\frac{dx}{\sqrt{-\left\{ \left[ x - \left( \frac{a + b}{2} \right) \right]^2 + \frac{4ab - a^2 - 2ab - b^2}{4} \right\} }}} \\ &= \int_a^b{\frac{dx}{\sqrt{- \left\{ \left[ x - \left(\frac{a + b}{2}\right) \right]^2 - \frac{a^2 - 2ab + b^2}{4} \right\} }}} \\ &= \int_a^b{ \frac{dx}{\sqrt{ - \left\{ \left[ x - \left(\frac{a + b}{2}\right) \right]^2 - \left(\frac{a - b}{2}\right)^2 \right\} }} } \\ &= \int_a^b{\frac{dx}{\sqrt{ \left( \frac{a - b}{2} \right)^2 - \left[ x - \left(\frac{a + b}{2}\right) \right]^2 }}} \end{align*} \]Calculate the following integral:
$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$
Yes, that's what I had assumed, but you're right, it's important to note thatThe most important thing in order to make this work: $b>a.$
if we miss this point ,we are going to do big mistakeThe most important thing in order to make this work: $b>a.$
That might seems a bit strange but this can be solved using the beta function !Calculate the following integral:
$$ \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}} $$