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Integration using Leibnitz theorem

  1. Sep 20, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Prove that $$\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx=\pi \arcsin(b)$$
    Where ##|b|\le1##

    2. Relevant equations
    $$\frac{d}{dx}\big(\int_a^bf(t)dt\big)=\int_a^b\frac{\partial f(t)}{\partial x}dx$$

    3. The attempt at a solution
    Let $$f(b)=\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx$$
    Using Leibnitz theorem, $$f'(b)=\int_{-\pi/2}^{\pi/2}\frac{1}{1+b\sin x}dx$$
    Now, ##\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}##
    After simplifying and substituting ##\tan(x/2)=u##, I got
    $$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$
    Now, for ##b=0##, its clear that ##f(b)=0##.
    So $$f(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$
    After putting ##b=\cos\phi## since ##|b|\le 1##, I got ##f(b)=\frac{\pi}{\sin({\arccos(b)})}## which is wrong.
    Also, the new ##f(b)## does not satisfy ##f(0)=0##

    Is my method correct?
     
    Last edited: Sep 20, 2015
  2. jcsd
  3. Sep 20, 2015 #2

    andrewkirk

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    How did you get this:

    $$
    f'(b)=\int_{-\pi/2}^{\pi/2}\frac{1}{1+b\sin x}dx
    \ \ ?$$

    That's not what I get when I take ##\frac{\partial}{\partial b}## of the integrand in ##f(b)## .
     
  4. Sep 20, 2015 #3

    Titan97

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    Sorry. Its ##log(1+b\sin x)##. I have edited the question.
     
  5. Sep 22, 2015 #4
    You made a mistake.
    $$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big] $$
    After this is your mistake.
    $$ \rightarrow \ f(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$

    It is $$f'(b)$$ and NOT $$f(b)$$

    Okay I think I started nitpicking, I am sorry. I think you messed up the last Integration part.

    $$ \int f'(b) db = \int ((\frac{2}{\sin \phi})\arctan(\tan (\frac{\pi}{2}- \phi) + \arctan (\tan \phi)) \times (-sin \phi) d(\phi)$$

    Integrate. Put f(0) = 0.
     
    Last edited: Sep 22, 2015
  6. Sep 22, 2015 #5

    Titan97

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    Thank you for pointing out the error. So,
    $$f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}$$
    Integrating ##d(f(\cos\phi))=g(\phi)d\phi##,
    $$f(\cos\phi)=\pi\log\tan\frac{\phi}{2}+C$$
    $$f(b)=\pi\log\sqrt{\frac{1-b}{1+b}}$$
    Now I get a different answer.
     
  7. Sep 23, 2015 #6
    Don't put $$\arctan (tan (\frac{\pi}{2} - \phi))$$ as $$(\arctan\cot\phi)$$ I wrote $$\arctan (tan (\frac{\pi}{2} - \phi))$$ deliberately, cause I have some devious plans for it. Apply some manipulation



    (like taking out ph...):wink:
     
  8. Sep 23, 2015 #7

    Titan97

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    When I substituted ##b=cos\phi##, I got ##\arctan\cot\phi## and not ##\arctan\tan(\frac{\pi}{2}-\phi)##
    Then ##\arctan x= \frac{\pi}{2}-\cot^{-1}x##.
     
  9. Sep 23, 2015 #8
    I see the problem there.
    1. Make a right angled triangle, take an angle as phi.
    2. Since sum of angles of triangle is pi hence the other angle has to be pi/2 - phi.
    3. Find tan of phi is perpendicular upon base [Let's say = AB/BC, ABC is triangle and B is right angle and C is phi].
    4. cot of (pi/2 -phi) is base upon perpendicular [From above we get cot (pi/2-phi) = AB/BC]
    5. From point 3. and 4. we have tan (phi) = cot (pi/2 -phi).
    6. I leave to you to prove cot (phi) = tan (pi/2 -phi).
    So even if you are ready to accept point 6 without proving it, you get, the manipulation I did. Can you get the answer from where I left? It is almost done, I am surprised that you did it so far and something so trivial is stopping from getting the answer. I think you should revise some basic trigonometry.
     
  10. Sep 23, 2015 #9

    Titan97

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    Proof:
    ##tan\phi=x##
    ##cot(\frac{\pi}{2}-\phi)=tan\phi=x##
    ##\frac{\pi}{2}-\phi=\cot^{-1}x##
    ##\phi=\frac{\pi}{2}-\cot^{-1}x##
    ##\tan^{-1}x=\frac{\pi}{2}-\cot^{-1}x##
     
  11. Sep 23, 2015 #10
    Why did you put that proof here? Are you even reading what I am writing? Look I don't understand what are you trying to say. If your confusion is cleared then good, if it is still there, then I am going to ask you to elaborate it, clearly. And by the way why did you put that proof here?
     
    Last edited: Sep 23, 2015
  12. Sep 23, 2015 #11
    I hope you do know $$\tan^{-1} (cot \phi) = \tan^{-1}(tan (\frac{\pi}{2}-\phi))= \frac{\pi}{2}-\phi$$
     
    Last edited: Sep 23, 2015
  13. Sep 23, 2015 #12

    Titan97

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    This is my confusion:
    $$f'(\cos\phi)=\frac{\pi}{\sin\phi}$$
    Integrating with respect to ##d\phi## I get wrong answer. But in the quoted message, why did you take ##f'(b)## when the function is ##f'(\cos\phi)##?
    Can't you integrate directly?
     
  14. Sep 24, 2015 #13
    Finally, deciphered. Yet another mistake in calculation and hopefully this will be my last thread to this post.
    Mistake -1. ##f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}##. Since R.H.S. is correct so you made a typo on computer here (started nitpicking again, sorry :biggrin:) It is ##f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi+\arctan\tan\phi)=\frac{\pi}{\sin\phi}## Notice the plus sign after tan inverse of cot of phi.

    Mistake -2(Actual one)
    I was typing while you posted this message but thanks to it I can clearly tell you where you made the mistake, otherwise it was getting very hard to do so with
    You can do it with cos as function too
    Note: In the quote you have written incorrectly ##f'(\cos\phi)=\frac{\pi}{\sin\phi}d\phi## no ##d\phi##
    $$f'(\cos\phi)=\frac{\pi}{\sin\phi}$$
    $$\rightarrow f'(\cos\phi)d(cos\phi)=\frac{\pi}{\sin\phi}d(cos\phi)$$
    $$\rightarrow \int f'(\cos\phi)d(cos\phi)= \int \frac{\pi}{\sin\phi}d(cos\phi)$$
    $$\rightarrow \int f'(\cos\phi)d(cos\phi)= \int \frac{\pi}{\sin\phi} \times (-\sin\phi)d(\phi)$$
    $$\rightarrow f(cos\phi) = -\pi (\phi) +c$$
    $$\rightarrow f(b)=-\pi (\cos^{-1}b) +c$$
    $$\rightarrow f(b)=-\pi (\frac {\pi}{2} - \sin ^{-1}b) +c$$
    $$\rightarrow f(b) =-\frac {\pi^2}{2}+ (\pi\sin^{-1}b) +c$$ and
    ##f(0)=0##


    But there is more to it. We both made a common mistake which does not interfere with answer.
    We both got $$\arctan cot\phi + \arctan tan\phi$$ It should be $$\arctan cot\frac{\phi}{2} + \arctan tan\frac{\phi}{2}$$.
    Though summation remains same but a we will have to accept that we got lucky and in some other place in some other question we will face the consequence if this mistake is not eradicated. We both gotta work on it.

    I hope I answered your question, so please this question Why were you confused that I wrote $$\arctan cot\phi + \arctan tan\phi = \arctan tan (\frac{\pi}{2} - \phi)+ \arctan tan\phi$$.

    And how did you quote me your last reply, you have quoted me entirely wrong. It was 2/sin(##\phi##) not 2sin(##\phi##) same with (##\pi##)/2
     
  15. Sep 24, 2015 #14

    Titan97

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    I wrote ##\phi/2## in notebook. I was typing in early morning just after I woke up. It was basically a communication problem.
    The answer given in the textbook however is ##\pi\sin^{-1}b##.
    But I know how to solve it now.
     
  16. Sep 24, 2015 #15
    Then why did you bother me with all the latex. I wasted 1.5 hours
     
  17. Sep 24, 2015 #16

    Titan97

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    You could have avoided LaTeX if you wanted to. I never forced you. Besides, the answer is still wrong.
     
  18. Sep 24, 2015 #17
    Why don't you tell me where
     
  19. Sep 24, 2015 #18

    Titan97

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    $$f(b)=\pi\sin^{-1}b$$
    That's the answer given in the text book.
     
  20. Sep 24, 2015 #19
    Put the damn value of c to get that. f(0) = 0
     
  21. Sep 24, 2015 #20
    c = (pi^2)/2
     
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