Asked by Anonymous
Does anybody has any idea how does 2(tan( θ)-θ)+C become 2[((√x^2-4)/2)-sec^-1*(x/2)]+C
Answers
Answered by
Steve
no idea, but the derivative of arcsec(x/2) is
2/(x√(x^2-4))
If secθ = x/2 then
tanθ = √(x^2-4)/2
and the derivative of
2(tanθ-θ) = 2(tan(arcsec(x/2))-arcsec(x/2)) is
2[x/(2√(x^2-4)) - 2/(x√(x^2-4))]
See whether that helps at all.
2/(x√(x^2-4))
If secθ = x/2 then
tanθ = √(x^2-4)/2
and the derivative of
2(tanθ-θ) = 2(tan(arcsec(x/2))-arcsec(x/2)) is
2[x/(2√(x^2-4)) - 2/(x√(x^2-4))]
See whether that helps at all.
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