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Integration, trig with double substitution

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Integral ( cos(x)/(1+cos(x))^.5 dx)

    2. Relevant equations

    3. The attempt at a solution

    Integral ( cos(x)/(1+cos(x))^.5 dx)

    Square it
    Integral ( cos(x)^2/(1+cos(x) dx)

    Multiply by the Conjugate
    Integral ( cos(x)^2/(1+cos(x) * (1 - cos(x))/(1 - cos(x)) dx)

    Take the square root and then multiply by sin(x)/sin(x)
    Integral ( cos(x)(1-cos(x))^.5/(sin(x) * sin(x)/sin(x) dx )

    Convert sin(x)^2 in denom to 1-cos(x)^2 then use u-sub, u= cos(x) du=-sin(x)
    Integral -( u(1-u)^.5/(1-u^2) du )

    w substitution w= (1-u)^.5 u = 1-w^2 du = -2w
    Integral (1-w^2)w^2/(1-(1-2w^2+w^4) dw)

    Reduce down to
    Integral 2((1-w^2)/(2-w^2) dw )

    I have more done but am leaving to go out to dinner now. Will post more of my work later if needed. Thanks in advance.
  2. jcsd
  3. Mar 17, 2008 #2


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    First, you can't 'square' an integral. Second, there is a nice trig identity you can use for the expression inside the square root. Look under half or double angle formulas. Can you find it?
  4. Mar 18, 2008 #3
    Okay, after you told me my first step was wrong, I started over. I looked at the half angle double angle stuff and didn't like how it was looking so I tried a similar approach to my first but fixed the first step.

    My solution:

    Integral ( cos(x)/(1+cos(x))^.5 dx)

    Conjugate/trig identities
    Integral ( cos(x)/(1+cos(x))^.5 (1-cos(x))^.5/(1-cos(x)^.5) dx)
    Integral ( cos(x)(1-cos(x))^.5)/(1-cos(x)^2)^.5 dx)
    Integral ( cos(x)(1-cos(x))^.5)/sin(x) dx)

    Times sin(x)/sin(x) and trig ident
    Integral( sin(x) * cos(x)(1-cos(x))^.5 / (sin(x)^2) dx)
    Integral( sin(x) * cos(x)(1-cos(x))^.5 / (1-cos(x)^2 dx)

    u-sub u=cos(x) du=-sin(x)
    Integral -( u(1-u)^.5/(1-u^2) du)

    w-sub w=(1-u)^.5 u=1-w^2 du=-2w dw
    Integral 2( (1-w^2)w^2/(1-w^2)^2 dw)

    Integral 2((1-w^2)/(2-w^2)

    Integral 2((1 - (1/(-w^2+2)) dw)

    Partial Fraction Decomposition
    2 * Integral (1dw) + .5(2)^.5 * Integral (1/(w-2^.5) dw) - .5(2)^.5 * Integral (1/(w+2^.5) dw)


    2w + .5(2)^.5*ln(w-2^.5) - .5(2)^.5*ln(w+2^.5) + C

    Resubstitute everything...
    2(1-cos(x))^.5 + .5(2)^.5*ln((1-cos(x))^.5 - 2^.5) - .5(2)^.5*ln((1-cos(x))^.5 + 2^.5) + C

    Fairly certain my steps are correct :D. Thanks for the assistance.

    Edit: Really need to figure out formatting lol.
  5. Mar 18, 2008 #4

    Gib Z

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    The Identities that Dick referred to where:

    [tex]cos (2t) = \cos^2 t - \sin^2 t = 2\cos^2 t -1[/tex].

    Both forms come in quite useful in evaluating this integral.
  6. Mar 18, 2008 #5
    I doubt this is much use but I was having a little play with that on my maths program and it came up with the rather horrendous answer:


    ??? hehe. It looked relatively simple?

    Have I expressed your question correctly?
  7. Mar 18, 2008 #6

    Gib Z

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    That expression can be simplified a whole heap!
  8. Mar 18, 2008 #7
    True but even so it's not very pretty.
  9. Mar 18, 2008 #8


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    … combine the ln()s … !

    If you type alt-v, it prints √ , which would help a lot :smile:
    erm … but you haven't finished! The examiners won't like you leaving two ln()s without combining them:

    ln(√(1-cos(x)) - √2) - ln(√(1-cos(x)) + √2)

    = ln((√1-cos(x)) - √2)/(√(1-cos(x)) + √2))

    = … ?
    (this can be very much simplified, and I think the result will look familiar to you! :smile:)
    Azureflames, you should do what Dick and Gib Z suggested - they're very wise, and their way only takes about five lines! :smile:
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