This is not the right place. However, I can't resist any potential test for senility.
Ok, after several false starts, and one look at
http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html, which I ended up not using, here goes. (Wolfram gave me the idea of trying to be more creative about the angle transforms, but I ended up not using theirs.
All the integrals here go from [0, pi/2]
1) 1/sqrt(sin(x)) dx
let (sin(y))^2=sin(x) (still range [0,1])
2) gives integrand 2dy cos(y) /cos(x) (check)
3) replacing cos(x) gives 2dy/sqrt(1+sin(y)^2)
4) can use cos(y) instead, same limits and form, just in opposite order,
2dy/sqrt(1+cos(y)^2)=
2dy/sqrt(2-sin(y)^2)=
sqrt(2)dy/sqrt(1-0.5sin(y)^2), still integrated [0,pi/2]
I guess we're done. Better check my work.
Mike W.
p.s. Here's another way:
Switch to 1/sqrt(cos(x)).
Substitute cos(x)=1-2sin(x/2)^2.
Switch to y=x/2 as the integration variable [0, pi/4]
Then I got stuck. ("Rusty" or "slowing" are much nicer words than "senile dementia".
Borrow trick from Wolfram: change variable to z via sin(z) = sqrt(2) sin(y).
Then crank through the usual manipulations, gives same result as above.
(published on 03/05/2011)
