Beta function proof

**simplependulum** · June 13th 2009, 11:34 PM

Prove that $\int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
where n is a real number ( guys you can prove whether n can be complex ... )

Then use this result to solve a Physics problem :
Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )

**Amer** · June 14th 2009, 12:32 AM

Originally Posted by simplependulum

Prove that $\int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
where n is a real number ( guys you can prove whether n can be complex ... )

Then use this result to solve a Physics problem :
Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )

you know that

$\beta (p,q) = 2\int_{0}^{\frac{\pi}{2}} sin^{2p-1}x cos^{2q-1}x dx$

you have

$\int_{0}^{\pi} sin^nx dx = \int_{0}^{\frac{\pi}{2}} sin^nx dx + \int_{\frac{\pi}{2}}^{\pi} sin^nx dx$

the first one equal

$\left(\frac{{\color{red}2}}{2}\right) {\color{red}\int_{0}^{\frac{\pi}{2}} sin^n x dx}=\frac{\beta(\frac{n+1}{2} , \frac{1}{2})}{2}$

since the red term equal beta and you have $2p-1 = n \Rightarrow p=\frac{n+1}{2}$ and $2q-1 = 0 \Rightarrow q=\frac{1}{2}$

the second integral

$\int_{\frac{\pi}{2}}^{\pi} sin^nx dx$ let $x=y+\frac{\pi}{2} \Rightarrow dx=dy$

$\int_{0}^{\frac{\pi}{2}} sin^n(y+\frac{\pi}{2}) dy =\int_{0}^{\frac{\pi}{2}} \left[siny(cos(\frac{\pi}{2}))+cosy(sin(\frac{\pi}{2}))\ right]^{n} dy$

$\Rightarrow \left(\frac{2}{2}\right)\int_{0}^{\frac{\pi}{2}} (cosy)^{n} dy =\frac{\beta(\frac{n+1}{2} , \frac{1}{2} )}{2}$

find the sum of the first and the second integarl you will have the answer

**Moo** · June 14th 2009, 12:34 AM

Hello,

Consider this formula of the beta function :

$\beta(x,y)=2\int_0^{\pi/2} \sin^{2x-1}\theta\cos^{2y-1}\theta ~d\theta$ (which you can get by a substitution $t=\sin^2\theta$ in the common formula of the beta function)

Then $\beta\left(\frac{n+1}{2},\frac 12\right)=2\int_0^{\pi/2} \sin^n\theta ~d\theta$

And since $\sin\left(\tfrac\pi 2-x\right)=\sin\left(\tfrac\pi 2+x\right) ~\dots$

**mr fantastic** · June 14th 2009, 03:03 AM

Originally Posted by simplependulum

Prove that $\int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
where n is a real number ( guys you can prove whether n can be complex ... )

Then use this result to solve a Physics problem :
Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )

Assuming the particle slides frictionlessly under gravity down the semicircle, the particle will leave the circle after the motion has gone through a polar angle equal to $\theta = \cos^{-1} \frac{2}{3}$ . So I'm not sure how the particle can end up at the bottom of the semicircle ....

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