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1

Here's some code that can solve this problem that I wrote a while back for another question. $TrigFns = {Sin, Cos, Tan, Csc, Sec, Cot}; (WRules = $TrigFns == (Through[$TrigFns[x]] /. x -> 2 ArcTan[t] // TrigExpand // Together) // Thread); invWRules = #[[1]] -> Solve[#, t, Reals] & /@ WRules; convert[expr_, (trig : Alternatives @@ ...


3

It is important to keep in mind that ArcTan[Tan[x]] is not always equal to x. When x is between -Pi/2 and Pi/2, it is valid. So we have In[24]:= ArcTan[Tan[x]] Out[24]= ArcTan[Tan[x]] And In[25]:= FullSimplify[ArcTan[Tan[x]], -Pi/2 < x < Pi/2] Out[25]= x Similarly, it follows that In[26]:= FullSimplify[ArcTan[Cos[x], Sin[x]], ...


2

the culprit here is AccuracyGoal: a = 3; c = 6; d = 0.00033; b = 2; NIntegrate[ x^3 (SphericalBesselJ[0, b x] + SphericalBesselJ[2, b x])/(4 d^2 x^2 + 9)^6 1/ 2 E^(-(1/2) x^2 - I x c - 1/2 a^2) (1 + E^(2 I x c) - I E^(2 I x c) Erfi[(x - I c)/Sqrt[2]] - I Erfi[(x + I c)/Sqrt[2]]), {x, 0, 8}, MaxRecursion -> 22, AccuracyGoal ...



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