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roadjester
 one year ago
Best ResponseYou've already chosen the best response.1who's the author of your book?

salparadise64
 one year ago
Best ResponseYou've already chosen the best response.0calculus early transcendental 7th E

roadjester
 one year ago
Best ResponseYou've already chosen the best response.1Damn, I've got Calculus 6th oh well

roadjester
 one year ago
Best ResponseYou've already chosen the best response.1Let me think; haven't done calc in a while

roadjester
 one year ago
Best ResponseYou've already chosen the best response.1I'm just gonna BS this, maybe something will come to me. \(\int{tan^nxdx=\int tan^{n1}}(x) tan(x)dx\)

salparadise64
 one year ago
Best ResponseYou've already chosen the best response.0pg 469 section 7.1 #53

salparadise64
 one year ago
Best ResponseYou've already chosen the best response.0yeah, i have the solution manual too, i was hoping someone would be able to explain it.

roadjester
 one year ago
Best ResponseYou've already chosen the best response.1oookkaay; I think the solution is selfexplanatory...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{}^{}\tan^n dx=\int\limits_{}^{}\tan^{n2}(x)\tan^2(x) dx=\int\limits_{}^{}\tan^{n2}(x)(\sec^2(x)1) dx\] \[=\int\limits_{}^{}\tan^{n2}(x)\sec^2(x)\int\limits_{}^{}\tan^{n2}(x) dx\] do a sub let u=tan(x) du=sec^2(x) dx and you will see you are almost done

salparadise64
 one year ago
Best ResponseYou've already chosen the best response.0@myininaya thank you!
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