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unimatix

  • one year ago

Integral 1/(x^2 sqrt(4-x^2)) dx

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  1. SolomonZelman
    • one year ago
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    \(x=2\sin\theta\)

  2. SolomonZelman
    • one year ago
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    @unimatix do you want a regular example where this method (trigonometric substitution) is used?

  3. unimatix
    • one year ago
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    That would be great honestly.

  4. SolomonZelman
    • one year ago
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    Ok, I am assuming you are familiar with a regular u-substitution, so I won't go to deep into explaining it.... \(\large\color{black}{ \displaystyle \int \frac{4}{\sqrt{1+x^2}}dx}\) \(\large\color{gray}{ \displaystyle x=\tan\theta}\) The derivative of \(\tan\theta\) is \(\sec^2\theta\), so your \(dx\) component will be: \(\large\color{gray}{ \displaystyle dx=\sec^2\theta{~} d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\left(\tan\theta\right)^2}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\tan^2\theta}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sec\theta}d\theta}\) \(\large\color{black}{ \displaystyle \int 4\sec\theta{~}d\theta}\) and then apply the formula for integral of secant

  5. SolomonZelman
    • one year ago
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    you will then substitute the x back in this example, but what I wanted to show is this mechanism of substituting a trignonometric function instead of x into the integral.

  6. SolomonZelman
    • one year ago
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    Do you want to come back to our original problem?

  7. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2\sqrt{4-x^2}}~dx}\) Try to do this substitution: \(\large \color{gray}{x=2\sin\theta}\) TELL ME: what is the derivative of \(2\sin\theta\) ?

  8. SolomonZelman
    • one year ago
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    (derivative with respect to \(\theta\) )

  9. unimatix
    • one year ago
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    I can follow what you did after you substituted in. But I can't understand what made you decide to do the trig substitution there.

  10. SolomonZelman
    • one year ago
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    What made me decide is that I want to make substitution in a way that the part in ssquare root containing x^2+C would become just a single trigonometric function inside the square root.

  11. unimatix
    • one year ago
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    Derivative should be 2cos right?

  12. SolomonZelman
    • one year ago
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    yes

  13. SolomonZelman
    • one year ago
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    2cos\(\theta\)

  14. unimatix
    • one year ago
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    Yeah that's what I meant, sorry

  15. unimatix
    • one year ago
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    so I set u = 2sin(theta) and du = 2cos(theta) dx right?

  16. SolomonZelman
    • one year ago
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    There is a rule for trigonometric substitution ... it will get intuitive after some time, but for now - the rule: ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: (\(a^2\) is just a constant) SUBSTITUTION \(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta \) \(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta \) \(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta \) ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼

  17. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\color{blue}{x}^2\sqrt{4-\color{blue}{x}^2}}~\color{royalblue }{dx}}\) Try to do this substitution: \(\large \color{blue}{x=2\sin\theta}\) \(\large \color{red }{dx=2\cos\theta ~d \theta }\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(\color{blue}{2\sin\theta}\right)^2\sqrt{4-\left(\color{blue}{2\sin\theta}\right)^2}}~\color{red }{2\cos\theta ~d \theta}}\)

  18. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\)

  19. SolomonZelman
    • one year ago
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    When you are done reading tell me whether you have followed everything so far, or there are so parts you don't get.

  20. unimatix
    • one year ago
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    Okay just worked through all the steps you went through

  21. SolomonZelman
    • one year ago
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    Ok, and now, you can tell this: \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4\left( 1-\sin^2\theta \right)}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\)

  22. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \cos\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\)

  23. SolomonZelman
    • one year ago
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    are you getting it?

  24. unimatix
    • one year ago
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    Yeah I get that.

  25. SolomonZelman
    • one year ago
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    Can you finish the problem up?

  26. SolomonZelman
    • one year ago
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    What function has a derivative of csc²(\(\theta\)) ?

  27. unimatix
    • one year ago
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    Is it cot(theta)?

  28. unimatix
    • one year ago
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    or I think -cot(theta)

  29. SolomonZelman
    • one year ago
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    YES, -cot(\(\theta\))

  30. unimatix
    • one year ago
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    -1/4 * cot(theta) ?

  31. unimatix
    • one year ago
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    - 1/(4 tan(theta)) ?

  32. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\csc^2\theta ~d \theta}\) \(\large\color{black}{\displaystyle-\frac{1}{4} \cot(\theta)+{\rm C} }\) and then you had that \(\large x=2\sin\theta \) so put back for x. \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\cos\theta}{\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\cos\theta}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\cos^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle\frac{-\sqrt{4-x^2}}{4x}+{\rm C} }\)

  33. SolomonZelman
    • one year ago
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    it was kind of long to put in for x.... And the check is: http://www.wolframalpha.com/input/?i=derivative+of+%28-%E2%88%9A%284-x%C2%B2%29%29%2F%284x%29

  34. unimatix
    • one year ago
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    Sorry it's taking me a little while to go through it.

  35. SolomonZelman
    • one year ago
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    I just rearranged the answer in terms of 2sin(\(\theta\)), and substituted for x.

  36. unimatix
    • one year ago
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    I'm confused as to how you got the third to last step

  37. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)

  38. SolomonZelman
    • one year ago
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    Are you referring to that?

  39. unimatix
    • one year ago
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    Yeah

  40. unimatix
    • one year ago
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    I think I've figured it out. That's tricky though.

  41. unimatix
    • one year ago
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    The last line is the final step right?

  42. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=}\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{blue}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1/4 \right){~}(4){~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{(1/4)}\cdot \sqrt{{~}4{~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{(1/2)\cdot \sqrt{{~}4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{{~}4-\left(2\sin\theta\right)^2}}{2\sin\theta}+{\rm C} }\)

  43. SolomonZelman
    • one year ago
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    By the way, I made a mistake: \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{red}{\displaystyle\frac{-\sqrt{4-x^2}}{8x}+{\rm C} }\)

  44. SolomonZelman
    • one year ago
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    the red is correct.

  45. unimatix
    • one year ago
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    Oh I should've seen that.

  46. unimatix
    • one year ago
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    |dw:1441678847661:dw|

  47. unimatix
    • one year ago
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    ahh the dx should be 2 cos(theta)

  48. SolomonZelman
    • one year ago
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    well, that is how I dx should be: `2cos(theta) d(theta)`

  49. SolomonZelman
    • one year ago
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    and that is exactly how I started to substitute \(2\sin\theta\) for \(x\).

  50. unimatix
    • one year ago
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    Oh sorry. I don't know what I was thinking . We just finished it.

  51. unimatix
    • one year ago
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    Thank you so much. I'm going to go try and work through it a second time on my own. THANK YOU. THANK YOU. THANK YOU.

  52. SolomonZelman
    • one year ago
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    Sure, remember the formas though ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: (\(a^2\) is just a constant) SUBSTITUTION \(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta\) \(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta\) \(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta\) ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ Like 4+x², is 2²+x² ---> form 1, x=atan(\(\theta\)).

  53. unimatix
    • one year ago
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    Will do!

  54. SolomonZelman
    • one year ago
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    Ok, yw!

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