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Math Help - MHF Integral Bee!

  1. #16
    Super Member Random Variable's Avatar
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    So who gets to post the next integral?
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  2. #17
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    Quote Originally Posted by Random Variable View Post
    So who gets to post the next integral?
    It is your turn , your answer is correct and you are the first person who answered the problem .
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  3. #18
    Super Member Random Variable's Avatar
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    \int^{2 \pi}_{0} \frac{\cos 2 \theta}{5-4 \cos \theta} \ d \theta
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  4. #19
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    Quote Originally Posted by Random Variable View Post
    \int^{2 \pi}_{0} \frac{\cos 2 \theta}{5-4 \cos \theta} \ d \theta
    \frac{\cos (2 \theta)}{5 - 4\cos \theta}=\frac{2 \cos^2 \theta - 1}{5 - 4 \cos \theta} = \frac{-1}{2} \cos \theta - \frac{5}{8} + \frac{17}{8} \cdot \frac{1}{5 - 4 \cos \theta}. we also have: J=\int_0^{2 \pi} \frac{d \theta}{5 - 4 \cos \theta} = 2 \int_0^{\pi} \frac{d \theta}{5 - 4 \cos \theta}. put \tan(\theta/2) = x to get: J=4\int_0^{\infty} \frac{dx}{1 + 9x^2} = \frac{2\pi}{3}.

    thus: \int_0^{2 \pi} \frac{\cos (2 \theta)}{5 - 4\cos \theta} \ d \theta=\frac{- 5 \pi}{4} + \frac{17 \pi}{12}=\frac{\pi}{6}.

    Problem: Prove that \int_0^{\frac{\pi}{2}} \frac{x}{1+\sin x} \ dx = \ln 2.
    Last edited by NonCommAlg; August 19th 2009 at 04:54 AM.
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  5. #20
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Problem: Prove that \int_0^{\frac{\pi}{2}} \frac{x}{1+\sin x} \ dx = \ln 2.

    \int \frac{x}{1+\sin x}dx

    \int \frac{x(1-sin x )}{1-sin^2 x}dx

    \int \frac{x}{cos^2 x}dx - \int \frac{x(sin x)}{cos^2 x }dx

    let x=u and dv=1/cos^2 x use integrate by parts for the left integral and for the right integral u=x and v=x(sin x)/cos^2 x

    but for the right integral you want to evaluate \int \frac{sin x}{cos^2 x} dx first , by sub u=cos x we can see that the integrate equal \frac{1}{cos x}

    x(tan x) - \int \frac{sin x}{cos x} dx -\left( x\left(\frac{1}{cos x}\right) - \int sec x dx\right)

    x(tan x) + ln (cos x) - \left ( \frac{x}{cos x} - ln (sec x + tan x ) \right)

    x(tan x) -\frac{x}{cos x} + ln (cos x ) + ln (sec x + tan x )

    \frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x)

    \frac{ x (sin x ) - x }{cos x} + ln ( 1+sin x) \mid ^{\frac{\pi}{2}}_{0}

     lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln (1 + sin \left(\frac{pi}{2}\right)- \frac{ 0 (0) - 0 }{1} - ln (1 + 0 )

     lim_{x\rightarrow \frac{\pi}{2}} \frac{x \sin x - x }{\cos x} + ln \left(1 + 1\right)- 0 - 0


    use lopital for the limit

    lim_{x\rightarrow \frac{\pi}{2}} \frac{x\cos x+(\sin x -1 ) }{-\sin x} + ln (2)

    the limit equal zero after the we sub pi/2 so the integral equal ln(2)

    here is my integral

    \int \ln (\sqrt{x^2 +1 } +1 ) dx
    Last edited by Amer; August 19th 2009 at 07:16 AM.
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  6. #21
    Senior Member pankaj's Avatar
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    I am deleting the post .I read the question wrongly
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  7. #22
    ynj
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    my answer is
    -\ln|\tan(\arctan(\frac{x}{2}))|+\frac{x}{\tan(\arc  tan(\frac{x}{2}))}+x\ln(\sqrt{x^2+1}+1)-\ln|x|+C
    the procedure is quite complex....
    it is too late to night, you may allow me showing it tomorrow..

    oh...it is not correct........sorry,there is a obvious mistake.(edited later..)
    Last edited by ynj; August 19th 2009 at 10:05 AM.
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  8. #23
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ynj View Post
    my answer is
    -\ln|\tan(\arctan(\frac{x}{2}))|+\frac{x}{\tan(\arc  tan(\frac{x}{2}))}+x\ln(\sqrt{x^2+1}+1)-\ln|x|+C
    the procedure is quite complex....
    it is too late to night, you may allow me showing it tomorrow..
    your solution is not correct ( or show me your work )

    another thing there is no boundaries for my integral
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  9. #24
    MHF Contributor red_dog's Avatar
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    \int\ln(\sqrt{x^2+1}+1)dx=\int x'\ln(\sqrt{x^2+1}+1)dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\frac{x^2}{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}}dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\left(1-\frac{1}{\sqrt{x^2+1}}\right)dx=

    =x\ln(\sqrt{x^2+1}+1)-x+\ln(x+\sqrt{x^2+1})+C
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  10. #25
    Super Member Random Variable's Avatar
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     \ln(x+\sqrt{x^{2}+1}) = \sinh^{-1}x
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  11. #26
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by red_dog View Post
    \int\ln(\sqrt{x^2+1}+1)dx=\int x'\ln(\sqrt{x^2+1}+1)dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\frac{x^2}{\sqrt{x^2+1}(\sqrt{x^2+1}+1)}dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}}dx=

    =x\ln(\sqrt{x^2+1}+1)-\int\left(1-\frac{1}{\sqrt{x^2+1}}\right)dx=

    =x\ln(\sqrt{x^2+1}+1)-x+\ln(x+\sqrt{x^2+1})+C
    post an integral
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  12. #27
    MHF Contributor red_dog's Avatar
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    1) Let f,g:[-a,a]\to\mathbb{R} two continuous functions, f odd, g even. Then

    \int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_0^a  g(x)dx

    2) Calculate \int_{-1}^1\frac{x^{2008}}{1+x^{2007}+\sqrt{1+x^{4014}}}d  x

    Edited.
    Last edited by red_dog; August 19th 2009 at 12:24 PM.
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  13. #28
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    Quote Originally Posted by red_dog View Post
    1) Let f,g:[-a,a]\to\mathbb{R} two continuous functions, f odd, g even. Then

    \int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_{-a}^af(x)dx
    for those who are working on this problem: there's a typo in the problem. the RHS should be \int_0^a g(x) \ dx instead of \int_{-a}^a f(x) \ dx.
    Last edited by NonCommAlg; August 19th 2009 at 11:49 AM.
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  14. #29
    Super Member PaulRS's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    for those who are working on this problem: there's a typo in the problem. the RHS should be \int_{-a}^a g(x) \ dx instead of \int_{-a}^a f(x) \ dx.
    It should be <br />
\int_{ - a}^a {\tfrac{{g\left( x \right)}}<br />
{{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx}  = \int_0^a {g\left( x \right) \cdot dx} <br />
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  15. #30
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    Quote Originally Posted by PaulRS View Post
    It should be <br />
\int_{ - a}^a {\tfrac{{g\left( x \right)}}<br />
{{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx} = \int_0^a {g\left( x \right) \cdot dx} <br />
    haha ... you're right. that's what i meant! i fixed it.
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