# Q & A: an elliptic integral

Q:
Im not sure if this is rite place or not to ask about math phys but i have a problem with it. I want to transform integral from 0 to pi/2 dx/(sqrt sin x) to one of forms for an elliptic integral of the first kind. Can anyone help? Thnx in advance
- Tom (age 18)
Surabaya, Indonesia
A:
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
, 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)

## Follow-Up #1: elliptic integral

Q:
Im sorry for checking late. Ok, i get you. So, we're letting sin x = sin^2 y. If i took the derivative of both side ill get cos x = d(u)^2/du dsiny/dy dy/dx (im letting sin y = u, then chain rule) = 2 siny cosy dy/dx. Thus dx = 2 siny cosy dy / cosx. Am i wrong? Then cosx replaced by what? I think the answer was right. But i need a little explanation about point 3) and 4)
- Tom (age 18)
INA
A:
step 3) integrand 2dy cos(y) /cos(x) = 2dy cos(y)/sqrt(1-sin^2(x))=
2dy cos(y)/sqrt(1-sin^4(y))=2dy cos(y)/sqrt(1-sin^2(y))sqrt(1+sin^2(y))=
2dy cos(y)/cos(y)sqrt(1+sin^2(y))=2dy/sqrt(1+sin(y)^2)

step 4) You see why you can integrate from pi/2 down to 0 instead, which is equivalent to using cos instead of sin from 0 to pi/2. So we have 2dy/sqrt(1+cos(y)^2)=
2dy/sqrt(2-sin(y)^2)=sqrt(2)dy/sqrt(1-0.5 sin(y)^2)

Mike W.

(published on 03/19/2011)