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ggrree

  • 2 years ago

Integral of tan^2(x)sec(x)

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  1. bahrom7893
    • 2 years ago
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    u is Sec(x)!!!!

  2. bahrom7893
    • 2 years ago
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    du is tan^2(x) dx.. the rest is for myin.

  3. TuringTest
    • 2 years ago
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    not quite bahrom

  4. myininaya
    • 2 years ago
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    \[\int\limits_{}^{}\tan(x) \cdot \tan(x) \sec(x) dx=\tan(x) \cdot \sec(x)-\int\limits_{}^{}\sec^2(x) \cdot \sec(x) dx\]

  5. bahrom7893
    • 2 years ago
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    oh wait lol i got it the other way around hahaha

  6. myininaya
    • 2 years ago
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    And you can look at the integral sec^3(x) and use integration by parts there

  7. myininaya
    • 2 years ago
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    oh wait i didn't need to do what i did

  8. myininaya
    • 2 years ago
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    \[\int\limits_{}^{}(\sec^2(x)-1)\sec(x) dx=\int\limits_{}^{}\sec^3(x) dx-\int\limits_{}^{}\sec(x) dx\]

  9. TuringTest
    • 2 years ago
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    that person is leaving, just so you know they told me on another post

  10. myininaya
    • 2 years ago
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    \[\int\limits_{}^{}\sec^3(x) dx=\int\limits_{}^{}\sec^2(x) \sec(x)dx=\tan(x) \sec(x)-\int\limits_{}^{}\tan(x) \sec(x) \tan(x) dx\] \[\tan(x) \sec(x)-\int\limits_{}^{}(\sec^2(x)-1) \sec(x) dx\]

  11. myininaya
    • 2 years ago
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    :(

  12. TuringTest
    • 2 years ago
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    I was doing problems with them for quite a while, but don't worry pretty sure they got what you were saying

  13. myininaya
    • 2 years ago
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    yay i think there is enough info here

  14. ggrree
    • 2 years ago
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    thanks to all of you! I got it now.

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