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purplemouse

  • 2 years ago

Integral sec t (tan t)^2 dt

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  1. Chlorophyll
    • 2 years ago
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    u = sect

  2. TuringTest
    • 2 years ago
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    hah, you got that from that sqrt(x-2)/sqrt(x-1) thing, right?

  3. TuringTest
    • 2 years ago
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    sect(tan^2t)=sect(sec^2t-1)=sec^3t-sect

  4. purplemouse
    • 2 years ago
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    I did

  5. TuringTest
    • 2 years ago
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    \[\int\sec^3 tdt\]\[u=\sec t\]\[dv=\sec^2 t\]integrate by parts

  6. purplemouse
    • 2 years ago
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    I've gotten that far in the equation already, just not sure what to do with the sec^3 t

  7. purplemouse
    • 2 years ago
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    ah, okay.

  8. TuringTest
    • 2 years ago
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    ^there it is this will be one of those integrals that repeats, so you will wind up adding the integral of sec^3 to both sides you will still have to integrate sec though, hope you remember how to do that ;)

  9. purplemouse
    • 2 years ago
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    I do :)

  10. sophierox
    • 2 years ago
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    correst

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