Is it possible to find the arclength of semicircle sqrt(16-x^2) w/o trig sub?

[RM_Mentock]

@miragesyzygy

it looks even messier in this textbox, but did I err somewhere, or does using trig sub (which we havn’t learned yet) make life a lot easier?

Have you been given a table, or access to a table, of integrals?

@Hari_Seldon

The OP is incoherent. Sqrt(16 - x^2) is not a circle; it is an expression. For example, when x = 1, it evaluates to sqrt(15) and so on.

But, f(x) = Sqrt(16 - x^2), or y = Sqrt(16 - x^2), are circles. That appears to be what the OP is about.