1. The problem statement, all variables and given/known data Arbitrary derivative of inverse trigonometric function: (sin-1x) = 1/(√1 - x2) 2. Relevant equations f-1(f(x)) = 1/f`(x) 3. The attempt at a solution So basically I learned about derivatives of trigonometric functions in class, and I thought maybe this would work: deriving the derivative of the regular sine function from its inverse sine, and then integrating that, to express a regular arbitrary trigonometric function in terms of x. Here is my thinking: (sin-1x) = 1/(√1-x2) f-1(sin x) = 1/(sin x)` = 1/(√1 - x2) Which is to say that: (sin x)` = (√1 - x2) Integrating this, we have sin x = ∫(1 - x2)½dx I haven't yet figured out how to evaluate this integral, but can someone tell me if my thinking is straight or not?