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Leibnitz rule

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

    solve ∫log(1+acosx) by differentiation under integral sign (limits are 0 to ∏)



    2. Relevant equations



    3. The attempt at a solution
    =∫(1/1+acosx)cosxdx(by leibinitz by differentiating partially WRT a.

    Then how do I proceed,can any one show me all the steps of reduction to standard integral forms??
     
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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    Welcome to PF; :)
    what is the base of the logarithm?
    by "acos(x)" do you mean ##a\cos(x)## or ##\arccos(x)##?
    No - because that would be against the rules. But we can try pointing you in promising looking directions in the hope you'll figure it out...

    I'd have been tempted to sub:
    ##e^u=1+\arccos(x) \Rightarrow x=\cos(1-e^u);##
    ... gets rid of both pesky functions at once.
    Then I'd look for a further substitution or explore integrating by parts a couple of times.

    The other way ... ##e^u=1+a\cos(x)## ;)

    hmmm... actually, Leibnitz's rule looks simpler... back in a tick.
     
    Last edited: Oct 11, 2012
  4. Oct 11, 2012 #3

    Simon Bridge

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    $$u(\alpha)=\int_0^\pi \ln|1+\alpha\cos(x)|dx = \int_0^\pi f(x,\alpha)dx= \int_0^\pi \frac{\partial}{\partial\alpha}f(x,\alpha)dx$$ ... note that ##\alpha\cos(\pi)=-\alpha## may be less than -1 making the argument of the logarithm negative.

    $$\frac{\partial}{\partial\alpha}f(x,\alpha)
    =\frac{\partial}{\partial\alpha}\ln|1+\alpha\cos(x)|=\frac{\cos(x)}{1+\alpha\cos(x)}$$... but would this have to be changed if ##|\alpha| > 1##

    so you end up with: $$u(\alpha)=\int_0^\pi \frac{\cos(x)}{1+\alpha\cos(x)}dx$$... which is where you are at.

    well I suppose you could get rid of the trig functions with ##z=\cos(x)## and the identity ##\sin(\arccos(z))=\sqrt{1-z^2}## but I don't hold out much hope.

    the other thing to try is ##\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})##
     
  5. Oct 11, 2012 #4

    Simon Bridge

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    I suppose there is always looking for p(x) and q(x) satisfying: $$q^2=1+\alpha\cos(x)$$ and $$q\frac{dp}{dx}-p\frac{dq}{dx}=\cos(x)$$... which will give you a first-order DE in p(x) to solve.

    Then, by the quotient rule: $$u(\alpha)=\frac{p}{q}$$... nothing looks tidy and a quick squiz at Wolfram's calculator suggests nothing will be.

    [note that ##1+\alpha\cos(x)## can be understood as ##|\vec{u}+\vec{v}|## where ##x## is the angle between the two vectors.]
     
    Last edited: Oct 11, 2012
  6. Oct 11, 2012 #5
    it is "acosx"..a is the parameter
     
  7. Oct 11, 2012 #6
    If I could get rid of cosx from numerator,then
    we could have substituted tan(x/2)=t
    dx=2dt/1+tsquare
    cosx=1-tsquare/1+tsquare.

    I tried dividing num&denom by cosx but no good.
     
  8. Oct 11, 2012 #7

    Simon Bridge

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    Ah - Weierstrass substitution is probably better earlier than that:$$\ln|1+\alpha\cos(x)| = \ln|1+t^2 -\alpha(1-t^2)|-\ln|1+t^2|$$... but I suspect we'll end up with the same trouble when we try to find f(t): dx=f(t)dt
    [edit]since the substitution is ##\tan(\frac{x}{2})=t## then $$dx = \frac{2}{1+t^2}dt$$... still not optimistic [/edit]

    I had another look at the substitution: ##e^z=1+\alpha\cos(x) \Rightarrow x=-\arccos(\frac{1-e^z}{\alpha})## because $$dx = -\frac{1}{\alpha}\sqrt{\frac{e^z}{2-e^z}}dz$$... or something.
    This would suggest an integration by parts to get rid of the stray z in the integrand, which may lead to hyperbolic functions.

    I think that's about the best I can do.
    The end result is very messy - the best you can do is try to ease the pain in getting there.

    Aside: looks unusually painful for a homework exercise - perhaps there is something in the context that will simplify the problem for you?
     
    Last edited: Oct 11, 2012
  9. Oct 11, 2012 #8
    simon,according to the exercise the final result does not contain any hyperbolic terms.
     
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