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Mathematics 6 Online
OpenStudy (anonymous):

need help proving reduction formula for calc

OpenStudy (anonymous):

please...

OpenStudy (anonymous):

@roadjester

OpenStudy (roadjester):

who's the author of your book?

OpenStudy (roadjester):

Stewart?

OpenStudy (anonymous):

james stewart

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

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...

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

OpenStudy (anonymous):

@myininaya thank you!

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