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Jnlucero

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  • one year ago
  • one year ago

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  1. Jnlucero
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    \[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]

    • one year ago
  2. Jnlucero
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    find the value.

    • one year ago
  3. lgbasallote
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    have you tried using completing the square method?

    • one year ago
  4. wio
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    Looks like a candidate for trig substitution.

    • one year ago
  5. Jnlucero
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    i do know the basic radical, but integrating radicals like that?.....oh thats confusing....

    • one year ago
  6. lgbasallote
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    you should use completing the square first....that would give you an integral that looks like \[\int \sqrt{x^2 - a^2}dx\] does that look familiar?

    • one year ago
  7. wio
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    Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).

    • one year ago
  8. wio
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    \[a=3,\ b=\sqrt{2}\]See what happens...

    • one year ago
  9. Jnlucero
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    then what will be next?

    • one year ago
  10. wio
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    Well if you did the substitution, you'd get \[ x=\frac{3}{\sqrt{2}}\tan(\theta)\quad dx = \frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \] \[ \Large \int_1^2 \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2}\frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \]

    • one year ago
  11. wio
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    Looks messy, but it really simplifies well.

    • one year ago
  12. Jnlucero
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    looks messy. then what will be next.....

    • one year ago
  13. wio
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    \[ \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)} \]

    • one year ago
  14. wio
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    \[ = \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta) \]

    • one year ago
  15. Jnlucero
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    what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?

    • one year ago
  16. wio
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    I was simplifying part of it. It's still there.

    • one year ago
  17. Jnlucero
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    ok......

    • one year ago
  18. Jnlucero
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    so how will i change it into \[\sec \theta d \theta\]

    • one year ago
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