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Beatles Group Title

\[\int\limits \frac{dx}{x \sqrt{81x ^{2}-16}}\]

  • 2 years ago
  • 2 years ago

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  1. agentx5 Group Title
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    Trig Sub! I know how to do these now! :-D One moment, I'll show you steps... I need to get my lunch from the microwave oven :-3

    • 2 years ago
  2. agentx5 Group Title
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    First while I go to get this food, 81 = 9^2 and 16 = 4^2

    • 2 years ago
  3. agentx5 Group Title
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    Trig sub. For this form with the 1/(x*sqrt(b^2*x^2-a^a)).... let \(x = \large\frac{4}{9}\sec\theta\) \(dx = \large\frac{4}{9}\tan\theta\sec\theta\) @Beatles watch what happens when you put this in place for "x" and "dx" :-)

    • 2 years ago
  4. agentx5 Group Title
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    \[\int\limits \frac{1}{x\cdot (9^2x^2-4^2)^{1/2}}\cdot dx\] \[\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]

    • 2 years ago
  5. agentx5 Group Title
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    Take the unit circle identity for sine and cosine, divide both side by cos^2 \[\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}\] Can you see what happens here, @Beatles? Can you write the identity here in terms of tangent-squared now?

    • 2 years ago
  6. agentx5 Group Title
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    Still there m8? :-D

    • 2 years ago
  7. experimentX Group Title
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    lol ... the fastest way ... do opposite.

    • 2 years ago
  8. Beatles Group Title
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    can we use the theorem of inverse trigo func. -_-

    • 2 years ago
  9. agentx5 Group Title
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    \[\int\limits \frac{1}{\sec\theta \cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\] @experimentX actually faster way is to use the integral tables ;-)

    • 2 years ago
  10. agentx5 Group Title
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    Err I mean wait that's 4/9 sec theta

    • 2 years ago
  11. agentx5 Group Title
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    Let me fix that...

    • 2 years ago
  12. experimentX Group Title
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    |dw:1342807447734:dw|

    • 2 years ago
  13. experimentX Group Title
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    |dw:1342807585448:dw|

    • 2 years ago
  14. experimentX Group Title
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    well ... note the pattern .. rest is just some scrupulous manipulation.

    • 2 years ago
  15. agentx5 Group Title
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    \[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot (9^2(\frac{4}{9}\sec\theta)^2-4^2)^{1/2}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\int\limits\limits \frac{1}{\frac{4}{9}\sec\theta\cdot \sqrt{16\tan^2\theta}}\cdot (\frac{4}{9}\tan\theta\sec\theta \ \ d\theta)\] \[\frac{1}{9} \int\limits\limits \frac{9}{4} d\theta\]

    • 2 years ago
  16. experimentX Group Title
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    still i recommend the trig substitution agentx5 did

    • 2 years ago
  17. agentx5 Group Title
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    Things cancel out, a lot. I get that on my scratchpad

    • 2 years ago
  18. agentx5 Group Title
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    |dw:1342807846986:dw|

    • 2 years ago
  19. agentx5 Group Title
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    SOH-CAH-TOA: SOH : \(\sin \theta = \large \frac{opposite}{hypotenuse} \) CAH : \(\cos \theta = \large \frac{adjacent}{hypotenuse} \) TOA : \(\tan \theta = \large \frac{opposite}{adjacent} \) \(\huge \frac{1}{\cos\theta} = \sec\theta\) Therefore: \(\huge \theta = sec^{^-1}(\frac{hyp}{adj})=sec^{-1}(\frac{9x}{4})\) Agreed @TuringTest & @experimentX , and do you follow what I'm doing here so far @Beatles?

    • 2 years ago
  20. agentx5 Group Title
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    We're going to use this to replace theta after you do that (now simple) integral above

    • 2 years ago
  21. experimentX Group Title
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    well ... i call that more intuitive and geometrical approach.

    • 2 years ago
  22. experimentX Group Title
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    |dw:1342808370968:dw| the answer would be |dw:1342808423489:dw| sorry ... i forgot chain rule.

    • 2 years ago
  23. agentx5 Group Title
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    \[\huge \frac{1}{9} \int\limits \frac{9}{4} d\theta = \frac{1}{\cancel{9}} \left[ \frac{\cancel{9}\theta}{4} \right] = \frac{\sec^{-1}(\frac{9 x}{4})}{4}+C\]

    • 2 years ago
  24. agentx5 Group Title
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    That should be the answer I think ^_^

    • 2 years ago
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