- #1

- 8

- 0

*dy/dx + sqrt( (1-y^2) / (1-x^2) ) = 0*

dy/(1-y^2) + dx/(1-x^2) = 0

sin -1 y + sin -1 x = c [1]

or, taking the sine of both sides

x (1-y^2)^1/2 + y (1-x^2)^1/2 = sin c [2]

dy/(1-y^2) + dx/(1-x^2) = 0

sin -1 y + sin -1 x = c [1]

or, taking the sine of both sides

x (1-y^2)^1/2 + y (1-x^2)^1/2 = sin c [2]

My response/confusion with this...

[1] Ok so I'm fine with this, standard integral

[2] ?Where on Earth did this come from

just by taking the sin of the equation

above.

Please help if you can.