purplemouse
simplify: tan^3(arcsec (x/2))
x= 2sec t
sqrt(x^2 4) = 2 tan t
t = arcsec (x/2)



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anonymous
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dw:1329633693726:dw

anonymous
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it is all in the triangle

anonymous
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there is a picture where the arc secant is x/2

anonymous
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find remaining side by pythagoras, get
\[\sqrt{x^24}\]

anonymous
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then find the tangent, then cube the result

dumbcow
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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^24)sqrt(x^24) / 8

purplemouse
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so I have (x^2 4)\[\sqrt{(x ^{2}}4)\]

purplemouse
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I get it now