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Integral sec t (tan t)^2 dt

Mathematics
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u = sect
hah, you got that from that sqrt(x-2)/sqrt(x-1) thing, right?
sect(tan^2t)=sect(sec^2t-1)=sec^3t-sect

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Other answers:

I did
\[\int\sec^3 tdt\]\[u=\sec t\]\[dv=\sec^2 t\]integrate by parts
I've gotten that far in the equation already, just not sure what to do with the sec^3 t
ah, okay.
^there it is this will be one of those integrals that repeats, so you will wind up adding the integral of sec^3 to both sides you will still have to integrate sec though, hope you remember how to do that ;)
I do :)
correst

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