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simplify: tan^3(arcsec (x/2)) x= 2sec t sqrt(x^2 -4) = 2 tan t t = arcsec (x/2)
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|dw:1329633693726:dw|
it is all in the triangle
there is a picture where the arc secant is x/2
find remaining side by pythagoras, get \[\sqrt{x^2-4}\]
then find the tangent, then cube the result
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you could also do a straight substitution tan = sqrt(sec^2 -1) tan = sqrt(x^2/4 -1) tan = sqrt(x^2 -4)/2 tan^3 = (x^2-4)sqrt(x^2-4) / 8
so I have (x^2 -4)\[\sqrt{(x ^{2}}-4)\]
I get it now
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