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Integration of inverse trigonometric functions

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data

    (x+2)dx/√(4x-x2)

    2. Relevant equations

    why was the -2 in -2(x-2) was ignored?

    3. The attempt at a solution

    so first i let u= 4x-x2
    then, du=4-2x
    = -2(x-2)

    so to get (x+2) i equate it to (x-2)+4

    so ...
    (x+2)dx/√(4x-x2) = (x-2)+4dx/√(4x-x2)

    = (x-2)dx/√(4x-x2) + 4dx/√(4x-x2)


    is the solution right?
    and why was the -2 in -2(x-2) was ignored?
     
    Last edited: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2

    SammyS

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    Hello delapcsoncruz. Welcome to PF !

    Check your differentiation.
    du = (4-8x)dx​

    As to your question: "why was the -2 in -2(x-2) was ignored?" : What are you referring to ?
     
  4. Jan 11, 2012 #3
    sorry typographical error.. the u= 4x-x2
    so du = 4-2x = -2(x-2)
     
  5. Jan 11, 2012 #4

    SammyS

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    Not quite right.

    du = -2(x-2)dx .

    You're doing the method of substitution incorrectly. At any rate, the substitution you're trying won't help much.

    Complete the square for -x2 + 4x .

    You should then be able to let u = x-2 .
     
  6. Jan 12, 2012 #5
    thank you...
    so (x+2)dx / (4x-x2) = (x+2)dx / (4-(x+2)2)

    i don't know what is the next thing to do.. please help..
     
  7. Jan 12, 2012 #6

    Curious3141

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    The completion of the square has not been done properly. [itex]\sqrt{4x-x^2} \neq \sqrt{4 - {(x+2)}^2}[/itex]. Remember that you have to get a [itex]-4x[/itex] term from the square expression to end up with a (positive) 4x at the end.
     
    Last edited: Jan 12, 2012
  8. Jan 12, 2012 #7
    thank you for the correction...
    sorry its (x+2)dx / 4 - (x-2)2

    so what i am going to do next?
     
  9. Jan 12, 2012 #8

    Curious3141

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    Ah good, you got it. You didn't need the hint I added in my edit.

    OK, now you can do the sub. that Sammy was suggesting, then separate the integrand into two simpler quotients added together (separate the sum in the numerator). After that, it should become quite easy to see.
     
  10. Jan 12, 2012 #9
    how can i do the substitution? can you give me a hint.
     
  11. Jan 12, 2012 #10

    Curious3141

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    Start by subbing u = (x-2).

    What does the numerator become in terms of u? Denominator? What about dx in terms of du?
     
  12. Jan 12, 2012 #11
    i solve it and this is my solution..

    ∫ (x+2)dx / sqrt (4x-x2)
    let u= 4x -x2
    du=-2(x-2)dx

    so now the numerator will become [(x-2)+4]dx

    = ∫ [(x-2)+4]dx / sqrt (4x-x2)

    = ∫ (x-2)dx / sqrt (4x-x2) + ∫ 4dx / sqrt (4x-x2)

    let u = 4x -x2
    du = -2(x-2)dx

    by completing the square : 4x - x2 = 4-(x-2)2

    = -1/2 ∫ du / u1/2 + 4∫ dx / sqrt 4 -(x-2)2
    = -u1/2 + 4∫ du / sqrt (a2 -u2)

    = -(4x -x2) + 4 Arcsin (x-2)/2 +c


    is the answer right?
     
  13. Jan 12, 2012 #12

    Curious3141

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    Your final answer is right (and you decided to take a slightly more complicated route, but that's OK).

    However, some of your notation needs cleaning up (it would be grounds for marking you wrong or subtracting marks). For example, in one line, "u" has two different definitions in the two integrals. And then "a" is brought in out of nowhere (you were probably referencing a standard form, but still, the "a" doesn't belong in the middle of your working).
     
  14. Jan 12, 2012 #13
    thank you very much! :smile: ok i will revise my solution to make it more understandable... thank you again..
     
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