Consider the shaded region bounded by $\sin x$, $\cos x$, and $\tan x$:

enter image description here

We can define this as an ImplicitRegion by:

\[ScriptCapitalR] = 
 ImplicitRegion[(0 < x < 1) \[And] (y < Cos[x]) \[And] 
     (y < Tan[x]) \[And] (y > Sin[x]),
  {x, y}]

However, RegionPlot[\[ScriptCapitalR]] fails to yield a figure after 15 minutes (v 11.3, MacOS).

However if I'm "smart" and put the bounds as $0<x< \pi/4$ I do get a plot.

Moreover, its area,


does not give an analytic solution (even after RootReduce, Simplify, etc.) even though an analytic form exists.

(One can get a numerical value through N@RegionMeasure[\[ScriptCapitalR]], but I seek the analytic solution.)

I've tried various forms based on RegionIntersection[] and such, without success.

Of course I can use traditional calculus through Integrate and finding intersection points, but I'd like to compute the area more directly.

How can I 1) plot the region and (more importantly) 2) compute the analytic area?

  • 1
    $\begingroup$ for (1) try RegionPlot[DiscretizeRegion@\[ScriptCapitalR]]? $\endgroup$
    – kglr
    Sep 15, 2021 at 0:33
  • 2
    $\begingroup$ and for (2) FullSimplify[ToRadicals@RegionMeasure@\[ScriptCapitalR]] gives 1/2 (-1 + 2 Sqrt[2] - Sqrt[5] + ArcCsch[2]) (version 11.3 windows) $\endgroup$
    – kglr
    Sep 15, 2021 at 0:37
  • $\begingroup$ Yep. (1) AND (2) work. Post and I'll accept. Thanks (again). $\endgroup$ Sep 15, 2021 at 0:37
  • 3
    $\begingroup$ 12.3.1 work fine. Or using Region[\[ScriptCapitalR]] $\endgroup$
    – cvgmt
    Sep 15, 2021 at 1:30
  • $\begingroup$ How is that region finite? given that the region as described repeats itself in every interval {2 n Pi, (2 n+1/4) Pi} for all n Integer? Also, why does Mathematica considers only the region for n=0? $\endgroup$
    – rhermans
    Sep 15, 2021 at 9:17

1 Answer 1


1. Process with DiscretizeRegion before feeding it to RegionPlot:

ℛ = ImplicitRegion[(0 < x < 1) ∧ (y < Cos[x]) ∧ (y < Tan[x]) ∧ (y > Sin[x]), {x, y}]

RegionPlot[DiscretizeRegion @ ℛ]

enter image description here

2. A composition of FullSimplify and RegionMeasure gives an exact result:

FullSimplify @ RegionMeasure @ ℛ
  1/2 (-1 + 2 Sqrt[2] - Sqrt[5] + ArcCsch[2])

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