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Math Help - Integration Problem

  1. #1
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    Integration Problem

    I would be very grateful if someone could help me with finding the integral for this little fellow...

    1/ (x*((1-x)^0.5))

    I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

    Thank you in advance!!
    Jessica.
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  2. #2
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    Quote Originally Posted by j_clough View Post
    I would be very grateful if someone could help me with finding the integral for this little fellow...

    1/ (x*((1-x)^0.5))

    I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

    Thank you in advance!!
    Jessica.
    \int \frac{1}{x \sqrt{1 - x}} \, dx = \ln |\sqrt{1 - x} - 1| - \ln |\sqrt{1 - x} + 1| + C

    I might have time later to give some details if no-one else has.
    Last edited by mr fantastic; September 10th 2008 at 05:20 AM. Reason: No particular reason
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  3. #3
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    It's a standard form.

    (\text{arcsech(x)})' = \frac{-1}{x\sqrt{1-x^2}}

    Notice that in your integral, it is actually:
    \frac{1}{x\sqrt{1-(\sqrt{x})^2}}
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  4. #4
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    Quote Originally Posted by j_clough View Post
    I would be very grateful if someone could help me with finding the integral for this little fellow...

    1/ (x*((1-x)^0.5))

    I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

    Thank you in advance!!
    Jessica.
    It's a standard solution but if you've got to show steps, you can do this. I've done this question and it's got a few steps so I'll explain instead of showing all steps. You can use substitution but not just one, it requires multiple substitution.

    \displaystyle\int \frac{1}{(x)(1-x)^{0.5}} \ \mathrm{d}x

    Substituting u= \frac{1}{(1-x)^{0.5}} and doing the sorting out will lead to...

    2 \displaystyle\int \frac{1}{\left(1-\frac{1}{u^2}\right)u^2} \ \mathrm{d}u

    Here, substituting v = \frac{1}{u^2} and doing the sorting out will lead to...

    -2 \displaystyle\int \frac{1}{1-v^2}\ \mathrm{d}v

    Now, just split into partial fraction and integrate. (You may require one more substitution when integrating a fraction).

    Remember to substitute all variables ( u,v ...)to get the final answer in terms of x. The answer will be \ln ((1 - x)^{0.5}-1) - \ln ((1 - x)^{0.5}+1) + c
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  5. #5
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    Quote Originally Posted by Air View Post
    It's a standard solution but if you've got to show steps, you can do this. I've done this question and it's got a few steps so I'll explain instead of showing all steps. You can use substitution but not just one, it requires multiple substitution.

    \displaystyle\int \frac{1}{(x)(1-x)^{0.5}} \ \mathrm{d}x

    Substituting u= \frac{1}{(1-x)^{0.5}} and doing the sorting out will lead to...

    2 \displaystyle\int \frac{1}{\left(1-\frac{1}{u^2}\right)u^2} \ \mathrm{d}u

    Here, substituting v = \frac{1}{u^2} and doing the sorting out will lead to...

    -2 \displaystyle\int \frac{1}{1-v^2}\ \mathrm{d}v

    Now, just split into partial fraction and integrate. (You may require one more substitution when integrating a fraction).

    Remember to substitute all variables ( u,v ...)to get the final answer in terms of x. The answer will be \ln ((1 - x)^{0.5}-1) - \ln ((1 - x)^{0.5}+1) + c
    The substitution 1 - x = u^2 is probably a little easier.

    The reciprocal substitution x = \frac{1}{u} can also be used and leads to the standard result quoted by chops.

    It's not difficult to show that the two answers are equivalent.

    Quote Originally Posted by Chop Suey View Post
    It's a standard form.

    (\text{arcsech(x)})' = \frac{-1}{x\sqrt{1-x^2}}

    Notice that in your integral, it is actually:

    \frac{1}{x\sqrt{1-(\sqrt{x})^2}}
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  6. #6
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    Ahh... je comprends! I'll remember this double substitution technique.

    Thank you very much for all your replies.

    Jessica.
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