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OpenStudy (anonymous):
OpenStudy (anonymous):
when i try plugging in the y and x values i get 0=5
OpenStudy (anonymous):
how do i remove the sec
OpenStudy (anonymous):
because i know this is right
OpenStudy (anonymous):
For your answer did you sub \[\sqrt{x^2-4} \]
x = 2 sec(t)
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OpenStudy (anonymous):
yeah and i integrated to get int (2tan(x)2sec(x)tan(x)dx/2sec(x)
OpenStudy (anonymous):
dx= 2secxtanxdx
OpenStudy (anonymous):
sqrt(x^2-4)=2tanx
OpenStudy (anonymous):
Give me a moment while I attempt to get what you got doing the same substitution.
OpenStudy (anonymous):
if it helps lol..
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OpenStudy (anonymous):
they just use different y and x values
OpenStudy (anonymous):
That exactly what I did but instead of Theta I used t to make it easier for me.
OpenStudy (anonymous):
so.. where do i go from here?
OpenStudy (anonymous):
You have the correct answer but are just trying to remove the inverse secant?
OpenStudy (anonymous):
x = 2 sec(t)
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OpenStudy (anonymous):
i can't have any trig function in the equation
OpenStudy (anonymous):
It doesn't make sense that you wouldn't have the sec^-1 there
OpenStudy (anonymous):
Sec(x/2)^-1 = cos(x/2)
OpenStudy (anonymous):
Using identities you can change that to a cosine but other than looking back at the original substitutions im not sure what other changes you can make?
OpenStudy (anonymous):
@drbgonzal did you end up figuring it out?
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OpenStudy (anonymous):
Maybe they want it completely simplified.|dw:1405803300788:dw|