//adsauto //google 4 //verification Bing //verification yadex Proof that the integral from a to b of 1/√(x-a)(b-x) dx =π by three method

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


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



We will Proof that  ba1(xa)(bx)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)=ππ1!=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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