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Integral of tan^2(x)sec(x)

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u is Sec(x)!!!!
du is tan^2(x) dx.. the rest is for myin.
not quite bahrom

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Other answers:

\[\int\limits_{}^{}\tan(x) \cdot \tan(x) \sec(x) dx=\tan(x) \cdot \sec(x)-\int\limits_{}^{}\sec^2(x) \cdot \sec(x) dx\]
oh wait lol i got it the other way around hahaha
And you can look at the integral sec^3(x) and use integration by parts there
oh wait i didn't need to do what i did
\[\int\limits_{}^{}(\sec^2(x)-1)\sec(x) dx=\int\limits_{}^{}\sec^3(x) dx-\int\limits_{}^{}\sec(x) dx\]
that person is leaving, just so you know they told me on another post
\[\int\limits_{}^{}\sec^3(x) dx=\int\limits_{}^{}\sec^2(x) \sec(x)dx=\tan(x) \sec(x)-\int\limits_{}^{}\tan(x) \sec(x) \tan(x) dx\] \[\tan(x) \sec(x)-\int\limits_{}^{}(\sec^2(x)-1) \sec(x) dx\]
I was doing problems with them for quite a while, but don't worry pretty sure they got what you were saying
yay i think there is enough info here
thanks to all of you! I got it now.

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