Question

Evaluate in terms of the gamma function
\[\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\text d x\]

Answer 1

@UnkleRhaukus , where is gamma ??

Answer 2

Isn't gamma function the one used for the factorial of non-integer values?

Answer 3

That is correct
http://en.wikipedia.org/wiki/Gamma_function

Answer 4

In the question though it think we rearrange the equation to a form of the Beta function http://en.wikipedia.org/wiki/Beta_function
and then Gamma
\[B(m,n)=\frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)}\]

Answer 5

I was thinking of expanding the integrand as a binomial series, but it's too complicated.

Answer 6

I have the answer if that helps

Answer 7

i ll just use my calculator lol

Answer 8

\[=\frac{1}{4}\sqrt \pi\left(\frac{\Gamma (1/4)}{\Gamma(3/4)}-4\frac{\Gamma(3/4)}{\Gamma(1/4)}\right)\]

Answer 9

Lemme try my idea.
\[\sqrt{\frac{1-x^2}{1+x^2}}=\left ( -1+\frac{2}{1+x^2}\right )^{1/2}\]
Okay, I don't get how it became that. =))

Answer 10

NB: \[\Gamma(\frac1 2)=\sqrt \pi\]

Answer 11

@UnkleRhaukus , thanks for explaining gamma function. It is a new term which I came across today. Would it be helpful to you if I were to integrate the expression you gave with respect to dx and give you the answer?

Answer 12

i dont think that is the right method because the definitions of the gamma function and the beta functions are both integrals,
but if the answer is particularly illumination i would be interested

Answer 13

i cant even see how you would be able to integrate it

Answer 14

let
1+x^2 = u, 2x dx = du
x = (u - 1)^(1/2)

1 - x^2 = -u+1

\[ \int_{1}^{0} \frac{(1 - u)^{1/2}}{u^{1/2}2(u-1)^{1/2}} du\]

Answer 15

bad ... stuck here!!!

Answer 16

\[x = (u - 1)^{1/2} \]\[x^2=(u-1)\]\[x^2+1=u\]
?\[1 - x^2 ≠ -u+1\]?

Answer 17

working backwards from the answer
\[=\frac{1}{4}B(1/2,1/4)-\frac{1}{2}B(-1/2,3/4)\]

Answer 18

let x^2 = tan^2A, 2xdx = 2 tanA sec^2AdA
dx = sec^2AdA

\[\int \sqrt{\cos^2A - \sin^2A} \sec^2 A \;dA = \int (\cos2A)^{1/2} (2\cos2A- 1)^{-1}dA ??? stuck again\]

Answer 19

wait that looks good, maybe

Answer 20

\[B(m,n)=2\int\limits_0^{\pi/2} \cos^{(2m-1)}\theta \sin^{(2n-1)}\theta\text d \theta\]

Answer 21

there is 2 in the middle!!

Answer 22

\[2\cos2A=1-2\sin^2x\]
double angle formula

Answer 23

i think we have all the pieces

Answer 24

@experimentX
what kind of substitution is this called again
i can remember doing this a few years ago
\[x^2 = \tan^2A\]