Inversing my function doesn't work and maybe simplification oddity

Questions
,
1

Hi

The function : InverseFunction[(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c]

: no solution .

I also tried with Mathematica :

InverseFunction[(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c]

gives

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 tanh^-1(((b-1) tan(x/2))/Sqrt[b^2-1]))/(b^2-1)^(3/2)+Subscript[c, 1]))/c]

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 tanh^-1(((b-1) tan(x/2))/Sqrt[b^2-1]))/(b^2-1)^(3/2)+d))/c]

gives

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 (b-1) tan(x/2) coth(x))/(b^2-1)^2+d))/c]

it’s simplified here ? Why not above ? So the same :
InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 (b-1) tan(x/2) coth(x))/(b^2-1)^2+d))/c] gives
same .

Putting this in WolframAlpha :

still no solution .

Integrate[(1/(1 + a Cos[x]))^2, x] in Mathematica and WolframAlpha

( " Standard computation time exceeded… " )

don’t give the maybe simplification .

I haven’t Wolfram|Alpha Pro .

You can see Solving Kepler's 1st law as a function of time • Physics Forums to understand where the function come from : the goal is to have the Kepler’s 1st law in function of time .

Thanks .

2

Analytically,it works only for simple functions.See below

    f[x_] := Log[x]; InverseFunction[(f[x] /. x -> #) &][x]
    f[x_] := Sin[x]; InverseFunction[(f[x] /. x -> #) &][x]
    f[x_] := Exp[x] - x; InverseFunction[(f[x] /. x -> #) &][x]
   f[x_] := Exp[x] - x^2; InverseFunction[(f[x] /. x -> #) &][x](*Can't find *)
   
 InverseFunction[((Integrate[(1/(1 + a Cos[x]))^2, x]) /. x -> #) &][x](*Can't find *)

In Mathematica,from help pages of InverseFunction:

As discussed in “Functions That Do Not Have Unique Values”, many
mathematical functions do not have unique inverses. In such cases,
InverseFunction[f] can represent only one of the possible inverses for
f. InverseFunction is generated by Solve when the option
InverseFunctions is set to Automatic or True

Numerically with Plot

  Show[{ContourPlot[Sin[x] == y, {y, -1, 1}, {x, -Pi, Pi}], 
    Plot[ArcSin[x], {x, -1, 1}, PlotStyle -> {Dashed, Red}]}](*Example with Sin[x]*)


 Int = Integrate[(1/(1 + a Cos[x]))^2, x] /. a -> 2(*For a=2*)
 ContourPlot[Int == y, {y, -1, 1}, {x, -Pi, Pi}](*Inverse of integral*)
3

y(x)=(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c , y(0)=0

So if we have y , how to find x ? Must compute it everytime ?

What about the maybe simplification ?

Also the derivative of antiderivative doesn’t give the original function (1/(1 + a Cos[x]))^2 , WolframAlpha and Mathematica gives
(a cos(x))/((a^2-1) (a cos(x)+1))+(a^2 sin^2(x))/((a^2-1) (a cos(x)+1)^2)-((a-1) sec^2(x/2) coth(x))/(a^2-1)^2+(2 (a-1) tan(x/2) csch^2(x))/(a^2-1)^2 .

4

y=Integrate[(a/(1 + b Cos[x]))^2/c, x]

So if we have y , a , b , c and the antiderivative’s constant how to find x ? Must compute it everytime ?

What about the maybe simplification ?

Also the derivative of antiderivative doesn’t give the original function (1/(1 + a Cos[x]))^2 , WolframAlpha and Mathematica gives
(a cos(x))/((a^2-1) (a cos(x)+1))+(a^2 sin^2(x))/((a^2-1) (a cos(x)+1)^2)-((a-1) sec^2(x/2) coth(x))/(a^2-1)^2+(2 (a-1) tan(x/2) csch^2(x))/(a^2-1)^2 .

5

func = a^2 (1/(1 + b Cos[x]))^2/c
int = Integrate[func, x]
((a^2 (-((2 ArcTanh[((-1 + b) Tan[x/2])/Sqrt[-1 + b^2]])/(-1 + b^2)^(
3/2)) + (b Sin)/((-1 + b^2) (1 + b Cos))))/c*)
D[int, x] // FullSimplify
(a^2/(c (1 + b Cos[x])^2))
func == D[int, x] // FullSimplify
(True)

The derivative of antiderivative gives the original function .

for y[0]= 0 then we have:

 Limit[int, x -> 0](*Then C = 0 *)
 y[x] == int // Expand
 (*y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[x])/(
   3 (1 + 2 Cos[x]))*)

Solve[y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[
   x])/(3 (1 + 2 Cos[x])), x](*Solve can' t solve because is a Transcendental equation .*)

We can solve only numerically automatic and view the solution into on the Graph:

  a = 1; b = 2; c = 1; 
   ContourPlot[y == int, {y, -1, 1}, {x, -Pi, Pi}, 
   Frame -> False, Axes -> True, AxesLabel -> Automatic]

More info for Transcendental equation see here.