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experimentX

  • 3 years ago

Integration:

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  1. experimentX
    • 3 years ago
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    the answer of the integration is \[ {1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t + {\sin (4t)\over 4}\right)\]

  2. experimentX
    • 3 years ago
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    On the other hand, Mathematica expresses solution as \[ \frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}} \]

  3. experimentX
    • 3 years ago
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    this is minus \[ {1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t - {\sin (4t)\over 4}\right) \]

  4. Callisto
    • 3 years ago
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    Hmm.... \[\int \sqrt{(a-x)(x-b)}dx\]Let \(x=acos^2t + bsin^2t\) \[dx = (-2a\sin t\cos t + 2b \sin t\cos t)dt = (b-a)sin2t dt\] Then, the integral becomes (a-x) = a - (acos^2 t + bsin^2 t) = a (sin^2 t) - b sin^2 t = (a-b) sin^2 t (x-b) = (acos^2t + bsin^2t) - b = acos^2 t - b (cos^2 t) = (a-b) cos^2t \[\int \sqrt{(a-b)sin^2t \times (a-b)cos^2t} (b-a)sin2tdt\]\[=\int [(a-b)sint cost] (b-a)sin2tdt\]\[=-\frac{1}{2}\int (b-a)^2sin^22tdt\] \[=-\frac{1}{2}(b-a)^2\int sin^2 2tdt\]\[=-\frac{1}{2}(b-a)^2\int \frac{1-cos4t}{2}dt\]\[=-\frac{1}{4}(b-a)^2 (t-\frac{1}{4}sin4t)+C\]Anything wrong here?

  5. experimentX
    • 3 years ago
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    looks like i made error

  6. experimentX
    • 3 years ago
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    yeah i made error ... - ( 1 - cos ..) = -

  7. experimentX
    • 3 years ago
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    If we put \[ \sin (t) = \sqrt{x-a\over b-a} \\ \cos (t) = \sqrt{b-x\over b-a} \\ t = \arctan \sqrt { \left( x-a \over b-x\right) } \]

  8. experimentX
    • 3 years ago
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    we get pretty close to |dw:1349076028513:dw|

  9. experimentX
    • 3 years ago
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    this is pretty close from solution of WA

  10. experimentX
    • 3 years ago
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    Cleaning up the solution i get \[ \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right) \over 8 \]

  11. Callisto
    • 3 years ago
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    Why should we express t in terms of arc tan?

  12. experimentX
    • 3 years ago
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    I don't know ... I just want match up our answer with WA

  13. Callisto
    • 3 years ago
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    Perhaps... That help get a prettier t... \[tan t = sint / cost = \sqrt{\frac{x-a}{b-a}}\] *Still working on that*

  14. experimentX
    • 3 years ago
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    Let's graph our solution

  15. experimentX
    • 3 years ago
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    something is missing

  16. experimentX
    • 3 years ago
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    \[ \tan t = \sin t / \cos t = \sqrt{\frac{a-x}{x-b}} \]

  17. Callisto
    • 3 years ago
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    :'( I got into trouble!!! \[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}}{b-a}(1-\frac{x-a}{b-a})]+C \]\[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}(b-x)}{(b-a)^2}]+C \]\[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})]+\frac{1}{4}\sqrt{(x-a)(b-x)}(b-x)+C \]Blah... Doesn't seem right :'(

  18. experimentX
    • 3 years ago
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    I almost got same result as mathematica except for this one \[ \left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \]

  19. experimentX
    • 3 years ago
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    \[ =-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}}{b-a}(1-2\frac{x-a}{b-a})]+C \] you missed 2 here.

  20. Callisto
    • 3 years ago
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    Oh!!! Let me redo it....

  21. experimentX
    • 3 years ago
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    also change that value inside arctan it should be \[ \sqrt{x-a \over b-x}\]

  22. Callisto
    • 3 years ago
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    Sorry.... typos!!

  23. Callisto
    • 3 years ago
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    Once again, it doesn't look great at all... \[-\frac{1}{4}(a-b)^2 [ tan^{-1} (\sqrt{\frac{x-a}{b-x}}) - \frac{\sqrt{(x-a)(b-x)}}{b-a}[1-2(\frac{x-a}{b-a})]] +C\]\[=-\frac{1}{4}(a-b)^2 [ tan^{-1} (\sqrt{\frac{x-a}{b-x}}) - \frac{\sqrt{(x-a)(b-x)}}{b-a}(\frac{a+b-2x}{b-a})] +C\]\[=-\frac{1}{4}(a-b)^2 tan^{-1} (\sqrt{\frac{x-a}{b-x}}) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C\]

  24. experimentX
    • 3 years ago
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    so far so good every part matches solution from WA except \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) \]

  25. Callisto
    • 3 years ago
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    What is that in WA?

  26. experimentX
    • 3 years ago
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    Wolf ...

  27. Callisto
    • 3 years ago
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    I meant what wolf showed you...

  28. experimentX
    • 3 years ago
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    yep!! http://www.wolframalpha.com/input/?i=Integrate+sqrt%28%28a-x%29%28x-b%29%29

  29. experimentX
    • 3 years ago
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    cancel those terms at the front

  30. experimentX
    • 3 years ago
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    \[ \frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}} \]

  31. experimentX
    • 3 years ago
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    Cancelling those first terms we get \[ {1 \over 8}\left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)\]

  32. experimentX
    • 3 years ago
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    it seems we are correct!! Wolf is pretty nasty!!

  33. experimentX
    • 3 years ago
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    We need to prove that \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) = \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \] this is not correct ... however I'll show you this is equivalent in integrals!!

  34. Callisto
    • 3 years ago
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    No... not really, the coefficient of arctan (simpler one) is -1/4, while the coefficient of arctan (wolf) is -1/8..

  35. experimentX
    • 3 years ago
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    |dw:1349079549133:dw|

  36. experimentX
    • 3 years ago
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    there is a formula that \( \arctan(x) + \arctan(1/x)= {\pi \over 2}\) |dw:1349079814250:dw|

  37. experimentX
    • 3 years ago
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    hence we have \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) = {\pi \over 2} - \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\] But this \( \pi \over 2 \) is constant and you can ignore it. Hence the result of the wolf follows.

  38. Callisto
    • 3 years ago
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    But what about the coefficient??

  39. experimentX
    • 3 years ago
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    \[ =-\frac{1}{4}(a-b)^2 \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C \\ -\frac{1}{4}(a-b)^2 {1\over 2 }\left( {\pi \over 2} - \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C\\ \frac{1}{8}(a-b)^2 \left( \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) \\ +C -{\pi \over 16 } (a-b)^2\\ \text{ Just let } C -{\pi \over 16 } (a-b)^2 = C_1 \text{ some constnat} \]

  40. experimentX
    • 3 years ago
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    Looks like i missed some - somewhere.

  41. Callisto
    • 3 years ago
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    Ehhhhh... I love my solution *v*

  42. experimentX
    • 3 years ago
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    yeah .. no need to go though rigorous way as Wolf.

  43. experimentX
    • 3 years ago
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    the above is just equally right!!

  44. Callisto
    • 3 years ago
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    It's always good know to how to get the wolf solution. Thanks !!!

  45. experimentX
    • 3 years ago
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    well ... remember this \( \arctan(x) + \arctan(1/x)= {\pi \over 2} \) and after integration you can manipulate your solution by adding or subtracting constants. expanple:- \( c + \sin^2 x\) and \( c - \cos ^2 x\) are equivalent solution.

  46. Callisto
    • 3 years ago
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    Why.......

  47. experimentX
    • 3 years ago
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    |dw:1349081258153:dw|

  48. experimentX
    • 3 years ago
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    this is typical when you integrate \[ \int \sin x \cos x \; dx\] one time let sin(x) = u ... another time let cos(x) = u ... another time put sin(x)cos(x) = 1/2 sin(2x)

  49. Callisto
    • 3 years ago
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    Eh?! three ways to do it?!

  50. experimentX
    • 3 years ago
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    if the difference between two solution is constant then they are equivalent. well you could put it up that way. but they were all equivalent.

  51. Callisto
    • 3 years ago
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    Hmm.. I remember someone told me that when I got a different answer from the textbook, it may not be the case that I did it wrong. sometimes, the answers may be different by some constants.. Perhaps, that's the case :\

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