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anonymous

  • 5 years ago

integrate dx/9+4x^2

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  1. anonymous
    • 5 years ago
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    Seems like you want a trig substitution here. Something like let \(2x = 3tan(\theta)\)

  2. anonymous
    • 5 years ago
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    polpak continue

  3. anonymous
    • 5 years ago
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    Well, what do you end up with using that substitution?

  4. anonymous
    • 5 years ago
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    \[ dx=3/2\sec ^{2}x thetad \theta\]

  5. anonymous
    • 5 years ago
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    Yeah, though you have a stray x in there I'm assuming is a typo.

  6. anonymous
    • 5 years ago
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    So now convert your integral into the new form with respect to theta

  7. anonymous
    • 5 years ago
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    please help idk

  8. anonymous
    • 5 years ago
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    Well, you have \[\int \frac{1}{9+4x^2}dx\] Since 2x = 3tan\(\theta\) What does that denominator become?

  9. anonymous
    • 5 years ago
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    \[9\sec ^{2}\theta\]

  10. anonymous
    • 5 years ago
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    \(9 + 9tan^2\theta\)

  11. anonymous
    • 5 years ago
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    Which yes, then becomes \(9sec^2\theta\)

  12. anonymous
    • 5 years ago
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    Ok, now you said dx = (3/2)\(sec^2\theta\ d\theta\)

  13. anonymous
    • 5 years ago
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    So we have \[\frac{3}{2}\int\frac{sec^2\theta}{9sec^2\theta}d\theta\]

  14. anonymous
    • 5 years ago
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    And I think you can take it from there.

  15. anonymous
    • 5 years ago
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    yea i got it thx.............

  16. anonymous
    • 5 years ago
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    polpak pls solve\[\int\limits \tan ^{3}x\]

  17. anonymous
    • 5 years ago
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    polpak i need ur help

  18. anonymous
    • 5 years ago
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    Hrm.. I don't recall the strategy for powers of tan I'm afraid. I'll see if I can re-derive it.

  19. anonymous
    • 5 years ago
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    ok

  20. anonymous
    • 5 years ago
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    integration by parts

  21. anonymous
    • 5 years ago
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    chaguanas i know it but solve it

  22. anonymous
    • 5 years ago
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    ¬_¬ \[\int \tan^3x = \int \tan x \ (\sec^2-1) = \int \tan x\ \sec^2x - \int \tan x \]

  23. anonymous
    • 5 years ago
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    INewton pls continues

  24. anonymous
    • 5 years ago
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    For the first one, you could use a substitution of note it is of the form \[f\ '(x) \cdot f \ (x) \] and try by recognition. \[ \int -\tan x = \int \frac{-\sin x}{\cos x}\] Which is of the form \[\frac{f \ '(x)}{f\ (x)} \] (think ln )

  25. anonymous
    • 5 years ago
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    INewton pls solve \[\int\limits \sqrt{9-4x ^{2}}/x\]

  26. anonymous
    • 5 years ago
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    Same type deal as the other trig sub you had before except instead of tan you let 2x = \(3 sec\theta\)

  27. anonymous
    • 5 years ago
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    Lets see if you can do it yourself this time.

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