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Math Help - evaluate the following definite integral

  1. #1
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    integral(((ln(1+bsinx))/(sinx)),x,-π/2,π/2)

    I hope you can understand my way of presenting the question.......
    Last edited by mr fantastic; July 1st 2009 at 01:53 AM. Reason: Merged posts
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  2. #2
    Senior Member pankaj's Avatar
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    Is this your question
    Evaluate:
    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{log_{e}(1+b\si  n x)}{\sin x}dx
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  3. #3
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    Hi

    Try to use LATEX code

    Do you mean

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx
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  4. #4
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    The answer is  \pi \sin^{-1}{b}  ~,~  |b| \leq 1 Isn't it ?
    Last edited by simplependulum; July 1st 2009 at 12:30 AM.
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  5. #5
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    Hello,

    \ln(1+b\sin x)=\sum_{n\geq 1} \frac{b^n \sin^n x}{n} (where |b|<1)

    So I=\int_{-\pi/2}^{\pi/2} \sum_{n\geq 1} b^n \cdot \frac{\sin^{n-1} x}{n} ~dx

    Inverting the sum and the integral (using the proper theorems), we have :

    I=\sum_{n\geq 1} \frac{b^n}{n} \int_{-\pi/2}^{\pi/2} \sin^{n-1} x ~dx=\sum_{n\geq 2}\frac{b^{n-1}}{n-1} \cdot J_n

    where J_n=\int_{-\pi/2}^{\pi/2} \sin^n x ~dx (known as Wallis integral in French..couldn't find it in English)

    By a common method ( \sin^n x=\sin^{n-2} x-\cos x \cdot \cos x\sin^{n-2} x) and an integration by parts, we get the recursive relation : J_n=J_{n-2}+\frac{1}{n-1} \cdot J_n

    \Rightarrow \boxed{nJ_n=(n-1)J_{n-2}}

    We can deduce this :

    \int_0^{\pi/2} \sin^n x~dx=\begin{cases} \frac{(2p)!}{2^{2p}(p!)^2}\cdot \frac \pi 2 \quad \text{if } n=2p \\ \frac{2^{2p} (p!)^2}{(2p+1)!} \quad \text{if } n=2p+1\end{cases}

    So if n is even, J_n=\frac{n!}{2^n \left[\left(\tfrac n2\right)!\right]^2} \cdot \pi

    And if n is odd, J_n=0

    So we have :

    I=\pi\sum_{k\geq 1} \frac{b^{2k-1}(2k)!}{(2k-1)2^{2k}(k!)^2}

    But hey, this is the power series for arcsin !!!

    So finally, \boxed{I=\pi \arcsin(b)} (I hope I didn't get wrong in the power series)
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  6. #6
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    Here's an alternative way.

    Let <br />
I(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx noting that I(0) = 0.

    Then I'(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{1}{1+b\sin x}\:dx.

    Introducing the usual trick x = 2 \,tan^{-1}z, then I'(b) = 2 \int_{-1}^1 \frac{1}{z^2 + 2bz + 1}\,dz which readily integrates. So

    I'(b) = \frac{2}{\sqrt{1-b^2}} \left(\tan^{-1} \frac{b+1}{\sqrt{1-b^2}} - \tan^{-1} \frac{b-1}{\sqrt{1-b^2}} \right) = \frac{2}{\sqrt{1-b^2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{1-b^2}}

    Integrating wrt b give I(b) = \pi \sin^{-1} b noting the constant of integration is zero due to the fact that I(0) = 0.

    Same answer but a little different approach.
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  7. #7
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    Quote Originally Posted by Danny View Post
    Here's an alternative way.

    Let <br />
I(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx noting that I(0) = 0.

    Then I'(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{1}{1+b\sin x}\:dx.

    Introducing the usual trick x = 2 \,tan^{-1}z, then I'(b) = 2 \int_{-1}^1 \frac{1}{z^2 + 2bz + 1}\,dz which readily integrates. So

    I'(b) = \frac{2}{\sqrt{1-b^2}} \left(\tan^{-1} \frac{b+1}{\sqrt{1-b^2}} - \tan^{-1} \frac{b-1}{\sqrt{1-b^2}} \right) = \frac{2}{\sqrt{1-b^2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{1-b^2}}

    Integrating wrt b give I(b) = \pi \sin^{-1} b noting the constant of integration is zero due to the fact that I(0) = 0.

    Same answer but a little different approach.
    hi sir,
    plz explain what I'(b) actually is and how do we differentiate a portion within the integration symbol with constant limits of integration........
    I'm a precollege student preparing for an entrance exam.We were not introduced to these type of problems,still I have got such a problem in a test.I hope sincerely you people will definitely help me out in these kind of problems
    thank you very much.
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  8. #8
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    Quote Originally Posted by Moo View Post
    Hello,

    \ln(1+b\sin x)=\sum_{n\geq 1} \frac{b^n \sin^n x}{n} (where |b|<1)

    So I=\int_{-\pi/2}^{\pi/2} \sum_{n\geq 1} b^n \cdot \frac{\sin^{n-1} x}{n} ~dx

    Inverting the sum and the integral (using the proper theorems), we have :

    I=\sum_{n\geq 1} \frac{b^n}{n} \int_{-\pi/2}^{\pi/2} \sin^{n-1} x ~dx=\sum_{n\geq 2}\frac{b^{n-1}}{n-1} \cdot J_n

    where J_n=\int_{-\pi/2}^{\pi/2} \sin^n x ~dx (known as Wallis integral in French..couldn't find it in English)

    By a common method ( \sin^n x=\sin^{n-2} x-\cos x \cdot \cos x\sin^{n-2} x) and an integration by parts, we get the recursive relation : J_n=J_{n-2}+\frac{1}{n-1} \cdot J_n

    \Rightarrow \boxed{nJ_n=(n-1)J_{n-2}}

    We can deduce this :

    \int_0^{\pi/2} \sin^n x~dx=\begin{cases} \frac{(2p)!}{2^{2p}(p!)^2}\cdot \frac \pi 2 \quad \text{if } n=2p \\ \frac{2^{2p} (p!)^2}{(2p+1)!} \quad \text{if } n=2p+1\end{cases}

    So if n is even, J_n=\frac{n!}{2^n \left[\left(\tfrac n2\right)!\right]^2} \cdot \pi

    And if n is odd, J_n=0

    So we have :

    I=\pi\sum_{k\geq 1} \frac{b^{2k-1}(2k)!}{(2k-1)2^{2k}(k!)^2}

    But hey, this is the power series for arcsin !!!

    So finally, \boxed{I=\pi \arcsin(b)} (I hope I didn't get wrong in the power series)
    sir can you plz explain me the first step??
    thanks in advance...........
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  9. #9
    Moo
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    Quote Originally Posted by IIT 2010 View Post
    hi sir,
    plz explain what I'(b) actually is and how do we differentiate a portion within the integration symbol with constant limits of integration........
    I'm a precollege student preparing for an entrance exam.We were not introduced to these type of problems,still I have got such a problem in a test.I hope sincerely you people will definitely help me out in these kind of problems
    thank you very much.
    http://en.wikipedia.org/wiki/Leibniz_integral_rule

    Quote Originally Posted by IIT 2010 View Post
    sir can you plz explain me the first step??
    thanks in advance...........
    Well, it's a basic power series (google for it)

    We know (geometric series) that if |y|<1, then \frac{1}{1+y}=\sum_{k\geq 0} y^k

    Integrate from 0 to x :
    \ln(1+x)=\int_0^x \frac{1}{1+y} ~dy=\sum_{k\geq 0} \frac{x^{k+1}}{k+1}=\sum_{k\geq 1}\frac{x^k}{k}
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