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

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### 1 Answer

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.**