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Specific treatment and solution of inverse function problem in advanced mathematics. Thank you.
(1)∫-3π≤x ≤- 2π, ∴- 1≤cosx≤ 1, that is,-1≤y≤ 1.

The inverse function of y = cosx is y=arccosx (- 1≤x≤ 1).

(2)∫y = arcsine √(2x+ 1)

∴siny=√(2x+ 1)≥0

And -π/2≤y≤π/2, ≤ 0 ≤ y ≤ π/2.

∴(siny)^2=2x+ 1

∴x= 1/2*(siny)^2- 1/2

That is, the inverse function is y =1/2 * (sinx) 2-1/2 (0 ≤ x ≤π/2).

Hope to adopt