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Math Help - Integration using... Inverse trig sub? or Solving the square?

  1. #1
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    Integration using... Inverse trig sub? or Solving the square?

    <br />
\int \sqrt{4 x-x^2} dx<br />
    Okay so I've ran into another bit of a snag...
    I know I can solve by the square and ill get
    sqrt(4-(x-2)^(2))

    but then again the question looks alot like the derivative of arcsin, which can be changed into arcsin...

    What would be best fastest most efficient way of getting there?
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  2. #2
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    Quote Originally Posted by Lolcats View Post
    <br />
\int \sqrt{4 x-x^2} dx<br />
    Okay so I've ran into another bit of a snag...
    I know I can solve by the square and ill get
    sqrt(4-(x-2)^(2))

    but then again the question looks alot like the derivative of arcsin, which can be changed into arcsin...

    What would be best fastest most efficient way of getting there?
    For:
    \int \sqrt{4-(x-2)^2} \,\ dx.
    Let x-2=2sin(\theta)d\theta.
    then dx = 2cos(\theta)d\theta.
    \int .... = \int \sqrt{4-4sin^2(\theta)} \,\ 2cos(\theta)d\theta
    =4\int cos^2(\theta)d\theta.
    Can you take it from here ?
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  3. #3
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    Yes. Thank you very much.
    However, I'm wondering when your doing that kind of substitution, how do you know that x-2=2sin(u) is equivalent to one another?

    I understand that sin(u) would equal x-2 / 2, but where would you get the 2 from to begin with?

    Do you assign 2 from x-2 as "a"?
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  4. #4
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    Quote Originally Posted by Lolcats View Post
    Yes. Thank you very much.
    However, I'm wondering when your doing that kind of substitution, how do you know that x-2=2sin(u) is equivalent to one another?

    I understand that sin(u) would equal x-2 / 2, but where would you get the 2 from to begin with?

    Do you assign 2 from x-2 as "a"?
    Do you the trigonometric substitution?
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  5. #5
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    Shouldn't the answer also be 2x+2sin2x?
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  6. #6
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    Somewhat, not really, the way my prof went over it just confused me, do you know of a link to a website thats really good at explaining it?
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  7. #7
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    Quote Originally Posted by Lolcats View Post
    Somewhat, not really, the way my prof went over it just confused me, do you know of a link to a website thats really good at explaining it?
    8. Integration by Trigonometric Substitution
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  8. #8
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    Wow, okay that makes a ton of sense. Thank you SO SO much.
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  9. #9
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    You say that it looked to you like "arcsine". Notice that you could have used either x-1= 2cos(u) or x-1= 2sin(u)- they both give the same result.

    You could also argue this way: Factor the "4" out: \sqrt{4- (x-1)^2}= 2\sqrt{1- \frac{(x-1)^2}{4}} which should remind you of \sqrt{1- cos^2(\theta)}= \sqrt{sin^2(\theta)}= sin(\theta) or \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta) so that you can use either \frac{x-1}{2}= sin(\theta) or \frac{x-1}{2}= cos(\theta) to clear the square root.
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