- #1
andyrk
- 658
- 5
Why is π/2∫3π/2[cos-1t].dt = π/2∫3π/2-1.dt?
[.] denotes the greatest integer function. Is it because [cos-1t] = -1 in t ∈ [π/2,3π/2)? But [cos-1t] = 0 for x = 3π/2. So we have one 0 along with all the -1's in (π/2,3π/2]. So how can we substitute [cos-1t] = -1 for every x in (π/2,3π/2] even though [cos-1t] = 0 at x = 3π/2?
Similarly, how can we substitute 3π/2∫2π[cos-1t].dt = 3π/2∫2π0.dt even though [cos-1t] = 1 at x = 2π?
[.] denotes the greatest integer function. Is it because [cos-1t] = -1 in t ∈ [π/2,3π/2)? But [cos-1t] = 0 for x = 3π/2. So we have one 0 along with all the -1's in (π/2,3π/2]. So how can we substitute [cos-1t] = -1 for every x in (π/2,3π/2] even though [cos-1t] = 0 at x = 3π/2?
Similarly, how can we substitute 3π/2∫2π[cos-1t].dt = 3π/2∫2π0.dt even though [cos-1t] = 1 at x = 2π?
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