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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
u is Sec(x)!!!!
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
du is tan^2(x) dx.. the rest is for myin.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
not quite bahrom
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\int\limits_{}^{}\tan(x) \cdot \tan(x) \sec(x) dx=\tan(x) \cdot \sec(x)\int\limits_{}^{}\sec^2(x) \cdot \sec(x) dx\]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
oh wait lol i got it the other way around hahaha
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
And you can look at the integral sec^3(x) and use integration by parts there
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
oh wait i didn't need to do what i did
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\int\limits_{}^{}(\sec^2(x)1)\sec(x) dx=\int\limits_{}^{}\sec^3(x) dx\int\limits_{}^{}\sec(x) dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
that person is leaving, just so you know they told me on another post
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\int\limits_{}^{}\sec^3(x) dx=\int\limits_{}^{}\sec^2(x) \sec(x)dx=\tan(x) \sec(x)\int\limits_{}^{}\tan(x) \sec(x) \tan(x) dx\] \[\tan(x) \sec(x)\int\limits_{}^{}(\sec^2(x)1) \sec(x) dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I was doing problems with them for quite a while, but don't worry pretty sure they got what you were saying
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
yay i think there is enough info here
 2 years ago

ggrree Group TitleBest ResponseYou've already chosen the best response.0
thanks to all of you! I got it now.
 2 years ago
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