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salparadise64 Group Title

need help proving reduction formula for calc

  • 6 months ago
  • 6 months ago

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  1. salparadise64 Group Title
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    please...

    • 6 months ago
  2. salparadise64 Group Title
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    @roadjester

    • 6 months ago
  3. roadjester Group Title
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    who's the author of your book?

    • 6 months ago
  4. roadjester Group Title
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    Stewart?

    • 6 months ago
  5. salparadise64 Group Title
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    james stewart

    • 6 months ago
  6. roadjester Group Title
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    What edition?

    • 6 months ago
  7. salparadise64 Group Title
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    calculus early transcendental 7th E

    • 6 months ago
  8. salparadise64 Group Title
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    @abb0t

    • 6 months ago
  9. roadjester Group Title
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    Damn, I've got Calculus 6th oh well

    • 6 months ago
  10. roadjester Group Title
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    Let me think; haven't done calc in a while

    • 6 months ago
  11. salparadise64 Group Title
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    hmmmmm

    • 6 months ago
  12. roadjester Group Title
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    I'm just gonna BS this, maybe something will come to me. \(\int{tan^nxdx=\int tan^{n-1}}(x) tan(x)dx\)

    • 6 months ago
  13. salparadise64 Group Title
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    pg 469 section 7.1 #53

    • 6 months ago
  14. salparadise64 Group Title
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    yeah, i have the solution manual too, i was hoping someone would be able to explain it.

    • 6 months ago
  15. roadjester Group Title
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    oookkaay; I think the solution is self-explanatory...

    • 6 months ago
  16. myininaya Group Title
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    \[\int\limits_{}^{}\tan^n dx=\int\limits_{}^{}\tan^{n-2}(x)\tan^2(x) dx=\int\limits_{}^{}\tan^{n-2}(x)(\sec^2(x)-1) dx\] \[=\int\limits_{}^{}\tan^{n-2}(x)\sec^2(x)-\int\limits_{}^{}\tan^{n-2}(x) dx\] do a sub let u=tan(x) du=sec^2(x) dx and you will see you are almost done

    • 6 months ago
  17. salparadise64 Group Title
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    @myininaya thank you!

    • 6 months ago
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