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Analysis And Algebra
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Proof that the integral from a to b of 1/√(x-a)(b-x) dx =π by three method

gimath


\int_{a}^{b} \frac{1}{\sqrt{(x-a)(b-x)}}dx



We will Proof that  ∫ba1(x−a)(b−x)‾‾‾‾‾‾‾‾‾‾‾‾‾√dx=π  by three method :


The first method:

let: x-a＝(b-a)y  ⇒ x＝(b-a)y + a ⇒  dx＝(b-a)dy

     b-x＝b-(b-a)y-a ⇒ b-x＝(b-a)(1-y)

     x＝a  ⇒    y＝o      ,     x＝b     ⇒     y＝1



\int_{a}^{b} \frac{1}{\sqrt{(x-a)(b-x)}}dx


   
\int_{0}^{1}y^{-\frac{1}{2}}\cdot(1-y)^{-\frac{1}{2}}dy=\beta(\frac{1}{2},\frac{1}{2})


=Γ(12)Γ12Γ(1)=π‾‾√⋅π‾‾√1!=π






The second method:

 let: x =a cos²Ө+b sin²Ө ⇒  x-a= (b-a) sin²Ө ⇒  b-x=(b-a) cos²Ө

 dx＝-2a sinӨ . cosӨ dӨ + 2b sinӨ . cosӨ dӨ ＝ (b-a)sin 2Ө dӨ

                x＝b     ⇒     Ө＝π/2    ,    x＝a   ⇒     Ө ＝0

Integration math


2\int_{0}^{\frac{\pi}{2}}d\theta=2[\theta]_0^\frac{\pi}{2}=2[\frac{\pi}{2-0}]=\pi

The third method:

let:   x-a＝y    ⇒   x＝y+a   ⇒   dx＝dy
        x＝b     ⇒  y＝b-a    ,    x＝a  ⇒     y＝0


\int_0^{b-a}y^{-{\textstyle\frac12}}\;\cdot{(b-a-y)}^{-\frac12}\;dy={(b-a)}^{\frac12+\frac12-1}\;\cdot\beta(\frac12,\frac12)


                         
=\frac{\Gamma({\displaystyle\frac12})\Gamma(\frac12)}{\Gamma(1)}=\frac{\sqrt\pi\cdot\sqrt\pi}{1!}=\pi









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