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OpenStudy (anonymous):
please...
OpenStudy (anonymous):
@roadjester
OpenStudy (roadjester):
who's the author of your book?
OpenStudy (roadjester):
Stewart?
OpenStudy (anonymous):
james stewart
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OpenStudy (roadjester):
What edition?
OpenStudy (anonymous):
calculus early transcendental 7th E
OpenStudy (anonymous):
@abb0t
OpenStudy (roadjester):
Damn, I've got Calculus 6th oh well
OpenStudy (roadjester):
Let me think; haven't done calc in a while
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OpenStudy (anonymous):
hmmmmm
OpenStudy (roadjester):
I'm just gonna BS this, maybe something will come to me.
\(\int{tan^nxdx=\int tan^{n-1}}(x) tan(x)dx\)
OpenStudy (anonymous):
pg 469 section 7.1 #53
OpenStudy (anonymous):
yeah, i have the solution manual too, i was hoping someone would be able to explain it.
OpenStudy (roadjester):
oookkaay; I think the solution is self-explanatory...
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myininaya (myininaya):
\[\int\limits_{}^{}\tan^n dx=\int\limits_{}^{}\tan^{n-2}(x)\tan^2(x) dx=\int\limits_{}^{}\tan^{n-2}(x)(\sec^2(x)-1) dx\]
\[=\int\limits_{}^{}\tan^{n-2}(x)\sec^2(x)-\int\limits_{}^{}\tan^{n-2}(x) dx\]
do a sub
let u=tan(x)
du=sec^2(x) dx
and you will see you are almost done