Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I prove that $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi?$$ I'm just wondering if LHS even equal to the RHS in the first place? Thanks for the help!

share|improve this question
4  
What if $a=b=0$? –  M.B. Sep 9 '13 at 16:03
4  
@M.B. Who cares? =) –  Pedro Tamaroff Sep 9 '13 at 22:00

3 Answers 3

up vote 30 down vote accepted

Hint: Substitute $x=a\cos^2\phi+b\sin^2\phi$.

(Quite an overkill solution ...)

share|improve this answer
5  
This is such a brilliant substitution! –  L. F. Sep 9 '13 at 16:06
    
Amazing, may I ask how you came up with this substitution? –  DepeHb Sep 9 '13 at 16:35
1  
I'm having two analysis class right now ... –  user67258 Sep 9 '13 at 16:38
1  
Certainly they've done you some good. (+1) –  Chris Sep 9 '13 at 17:33

Hint:$$\int_a^b{(u-a)^{x-1}(b-v)^{y-1}}=(b-a)^{x+y-1}\frac{\Gamma(x).\Gamma(y)}{\Gamma(x+y)}$$

$$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=(b-a)^{\frac12+\frac12-1}\frac{\Gamma(\frac12).\Gamma(\frac12)}{\Gamma(1)}=\pi$$

share|improve this answer

Here is another method using square completion: $$ \begin{align} \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}&=\int_a^b \frac{dx}{\sqrt{-\left(x^2-(a+b)x+ab\right)}}\\ &=\int_a^b \frac{dx}{\sqrt{\frac{1}{4}(a-b)^2-\left(x-\frac{b}{2}-\frac{a}{2}\right)^2}} \, \text{square completion}\\ &=\int_{\frac{a-b}{2}}^{\frac{b-a}{2}}\frac{2du}{\sqrt{(a-b)^2-4u^2}}, \,\text{substituing } u=x-b/2-a/2 \end{align} $$ Let $u=\frac{(a-b)}{2}\sin(v)$ to complete.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.