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Mathematics 20 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

|dw:1329633693726:dw|

OpenStudy (anonymous):

it is all in the triangle

OpenStudy (anonymous):

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

OpenStudy (anonymous):

find remaining side by pythagoras, get \[\sqrt{x^2-4}\]

OpenStudy (anonymous):

then find the tangent, then cube the result

OpenStudy (dumbcow):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I get it now

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