Hard Improper integral question

[grapz]

a and b are constants such that a < b. Find the value of [int] from a to b of dx / [ (x - a)(b - x) ] ^1/2

The answer is pi. I have no idea what to do and where to start

 

[soroban]

Hello, grapz!

$\displaystyle a\text{ and }b\text{ are constants such that: }a < b.$

$\displaystyle \text{Find the value of: }\;\int^b_a \frac{dx}{\sqrt{(x - a)(b - x)}}$

The answer is $\displaystyle \pi$

I have an approach . . . does anyone want to try it?


$\displaystyle \text{Under the radical we have: }\;-x^2+ax + bx -ab \;=\;-[x^2 - (a+b)x + ab]$

$\displaystyle \text{Complete the square: }\;-\left[x^2 - (a+b)x + \frac{(a+b)^2}{4} \:+\:ab - \frac{(a+b)^2}{4}\right]$

. . $\displaystyle = \;-\left[x^2-(a+b)x + \frac{(a+b)^2}{4}\right] + \left[\frac{(a+b)^2}{4} - ab\right] \;= \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2+2ab + b^2}{4} - ab\right]$

. . $\displaystyle = \;-\left[x - \frac{a+b}{2}\right]^2 + \left[\frac{a^2-2ab + b^2}{4}\right] \;=\;-\left(x - \frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2$

$\displaystyle \text{The integral becomes: }\;\int^b_a\frac{dx}{\sqrt{ (\frac{a-b}{2} )^2 - (x - \frac{a+b}{2} )^2}}$


\(\displaystyle \text{Now let: }\:x-\frac{a+b}{2} \;=\;\frac{a-b}{2}\sin\theta\quad\hdots\)


Try it . . . you're in for a pleasant surprise . . .

 

[grapz]

$\displaystyle \text{The integral becomes: }\;\int^b_a\frac{dx}{\sqrt{ (\frac{a-b}{2} )^2 - (x - \frac{a+b}{2} )^2}}$

Wow. That works out really nicely. That integral just becomes arcsin [( 2x - a - b) / ( b - a)] + c. Then solving for the improper integral is much easier. Thanks

 

[Dr. Flim-Flam]

Soroban, very good. However when I plugged in the original definite integral on my TI-89, I got undefined and

when I plugged it into Maple 8, I got a bunch of goobly-gook with imaginary numbers. Does this mean some integrals (gasp)

have to be done the old fashion way?

 

[Dr. Flim-Flam]

I get -Pi, not Pi.

 

[Dr. Flim-Flam]

Soroban, does arcsin(-1) = -Pi/2 or 3Pi/2? If the first then the answer is -Pi,

if the second then the answer is Pi.

 

[galactus]

The answer is indeed Pi. Yes, you get undefined running it through the calculator. This is an e;lliptic integral(or very close to it).

It appears to be similar to Gauss's famous $\displaystyle \int_{-\infty}^{\infty}\frac{1}{\sqrt{(x^{2}+a^{2})+(x^{2}+b^{2})}}dx=\frac{\pi}{agm(a,b)}$. Where agm(a,b) is the arithmetic-geometric mean.

From which one gets $\displaystyle \int_{-\infty}^{\infty}\frac{1}{x^{2}+(agm(a,b))^{2}}dx=\frac{\pi}{agm(a,b)}$

 

[Dr. Flim-Flam]

galactus, I think I found a contradiction on my TI-89. The answer comes out to be arcsin(-1)-arcsin(1) = -Pi

However if I plug constants into the original integral, a<b on my TI-89, I will get Pi. However when I punch in arcsin(-1) into my

trusty TI-89, I get -PI/2, not 3Pi/2. Hence the TI-89 uses 3PI/2 to give Pi to the definite integral but gives -pi/2 for

arcsin(-1).