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How to solve using Leibniz Rule

  1. Jun 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Screenshot_2015_06_09_12_12_31.png

    2. Relevant equations


    3. The attempt at a solution
    2015_06_10_07_15_35.jpg

    here is i am still stuck.
     
  2. jcsd
  3. Jun 9, 2015 #2

    Zondrina

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    I believe there are quite a few typos in the problem statement if you are intending to use Leibniz rule. I for one have never seen an ##\text{In}## function before.

    The Leibniz rule in one dimension would be:

    $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$

    EDIT: It would be more appropriate to call this the fundamental theorem.
     
    Last edited: Jun 9, 2015
  4. Jun 9, 2015 #3
    at first i am use substitution. some one said
    what am i do now?
     
  5. Jun 9, 2015 #4

    Zondrina

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  6. Jun 9, 2015 #5
    i know the Leibniz's rule. i solve many mathematical term . bt i cann't solve this math with Leibniz rule. i am still stuck
    here. here limit is constant so 2nd and 3rd term will be zero.
     
  7. Jun 10, 2015 #6
    any one solve this mathematical term and prove it??
     
  8. Jun 10, 2015 #7

    Ray Vickson

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    Your application of Leibnitz' rule is correct. You could try to change variables to ##u = \cos(x)## in your integral.
     
  9. Jun 10, 2015 #8
    See this, example 3.
    Now let's do that daunting looking integral. Like I said before, this is a good problem.
    First write ##\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx## for the original integral.
    However ##\int \frac{1}{1+\alpha cos(x)} dx## itself is not nice, in-fact it's pretty nasty. It will be an interesting exercise to do, but even after that there is a good bit to do for the original problem. So I would recommend you use Wolfram for that for now, and proceed.
     
  10. Jun 10, 2015 #9
    i could not solve this math above analysis . i am trying to my best. but i am fail every time. and again try.....

    please see this and give me more idea to solve
    proxy.php?image=http%3A%2F%2Fs10.postimg.org%2F73jxqf5uh%2FScreenshot_2015_06_09_12_12_31.png

    Screenshot_2015_06_10_15_36_42.png
     
  11. Jun 10, 2015 #10

    ehild

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    You can even cheat :) using wolframalpha.com.
    The substitution tan(x/2)=u works well if you have a rational expression of trigonometric functions. Try.
     
  12. Jun 10, 2015 #11

    ehild

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    The derivation of the OP is correct. Read Leibnitz Rule http://en.wikipedia.org/wiki/Leibniz_integral_rule
     
  13. Jun 10, 2015 #12
    In regards to post #9:- I'm not entirely sure what you're trying to do here. Did you have a look at the links I provided in post #8. I think your trying to come up with a value for ##\int_0^{\pi} \frac{1}{1+\alpha cos(x)} dx## for ##|\alpha|<1##using ##\int_0^{\pi} \frac{1}{\alpha-cos(x)} dx=\frac{\pi}{\sqrt{\alpha^2-1}}## for ##|\alpha|>1##, however that + sign in the original integral makes a good difference.
    Does no one read my posts :-)
     
  14. Jun 10, 2015 #13

    Ray Vickson

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    You can write your integral for ##f'(\alpha)## as ##\int_0^{\pi/2} + \int_{\pi/2}^{\pi}##, then change variables to ##x \leftarrow \pi -x## in the second integral, to get
    [tex] f'(\alpha) = \int_0^{\pi/2} \left[ \frac{\cos(x)}{1 + \alpha \cos(x)} - \frac{\cos(x)}{1 - \alpha \cos(x)} \right] \, dx \\
    = -\int_0^{\pi/2} \frac{2 \alpha \cos^2(x)}{1 - \alpha^2 \cos^2(x)} \, dx [/tex]
    Changing variables to ##\tan(x) = y## produces an integral in ##y## that can be split up into partial fractions, giving two well-known and easily do-able integrations.
     
  15. Jun 10, 2015 #14

    ehild

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    ##\int \frac{1}{1+\alpha cos(x)} dx## is not that nasty.
    Use the identity ##\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}##. Substitute u=tan(x/2), x=2arctan(u), ##dx=\frac{2}{1+u^2}du##. The integration limits become 0-->infinite.
     
  16. Jun 10, 2015 #15

    SammyS

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  17. Jun 11, 2015 #16
    Yeah, your right. I didn't actually do the integral, I only saw the answer and thought, this might take some time.
     
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