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KarlaKalurky

integrate:

  • one year ago
  • one year ago

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  1. KarlaKalurky
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    |dw:1352086212618:dw|

    • one year ago
  2. KarlaKalurky
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    let u=tan theta du= sec^2 d(theta)

    • one year ago
  3. KarlaKalurky
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    could it be: |dw:1352086779624:dw| ?

    • one year ago
  4. KingGeorge
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    Are you told to do a u-sub of \(u=\tan(\theta)\)?

    • one year ago
  5. KarlaKalurky
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    yes.,

    • one year ago
  6. KingGeorge
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    Well then, you get \[\int \tan(\theta)^2\sec(\theta)^4 d\theta\]With a u-sub of \(u=\tan(\theta)\), we have \(du=\sec(\theta)^2d\theta\), so it seems like we get\[\int u^2\sec(\theta)^2 du\]or\[\int u^2 \frac{du^2}{d\theta}\]

    • one year ago
  7. KingGeorge
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    However, if we start back at \(du=\sec(\theta)^2d\theta\), and take the derivative again, we get \[du^2=2\tan(\theta)\sec(\theta)^2d\theta^2.\]Honestly, I'm not sure where we're supposed to go with that u-sub.

    • one year ago
  8. KarlaKalurky
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    |dw:1352087930624:dw| ?

    • one year ago
  9. KingGeorge
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    What I seem to be getting, is this. I'm not quite sure what to do with it.\[\int u^2\frac{(du)^2}{d\theta}\]

    • one year ago
  10. KarlaKalurky
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    uhm...just try to write your solution. ill appreciate it :)

    • one year ago
  11. KingGeorge
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    Hold on a second. I think I know what you need to do. Let \(u=\tan(\theta)\). Then use the identity \(\sec(\theta)^2=\tan(\theta)^2+1\). That means we now have\[\large \int \tan(\theta)^2(\tan(\theta)^2+1)\sec(\theta)^2 d\theta\]Now you make the u-sub.

    • one year ago
  12. KingGeorge
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    We get \[\large \int u^2(u^2+1) \;du=\int u^4+u^2 \;du.\] This you should be able to integrate. Then just substitute \(u=\tan(\theta)\) back in, and you're good.

    • one year ago
  13. KarlaKalurky
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    thanks a lot!! :)

    • one year ago
  14. KarlaKalurky
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    |dw:1352125444501:dw| |dw:1352125609551:dw|

    • one year ago
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