Given dy/dx=1/square root(36-x^2). What is y ?

To determine the primitive function y, we'll have to
compute the indefinite integral of the function.

If dy/dx =
1/sqrt(36-x^2) => dy = dx/sqrt[(6)^2 - x^2]

We'll
integrate both sides:

Int dy = Int dx/sqrt[(6)^2 -
x^2]

We'll recognize the
formula:

Int dx/sqrt(a^2 - x^2) = arcsin (x/a) +
C

Let a = 6 => Int dx/sqrt[(6)^2 - x^2] = arcsin
(x/6) + C

The required function y, when dy/dx
= 1/sqrt(36-x^2), is: y = arcsin (x/6) + C.