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Math Help - interesting integral problem (from old post)

  1. #1
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    interesting integral problem (from old post)

    \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}

    This problem was found on an old post and i thought it was an interesting problem. a hint was given which said remember that tan x = cot(pi/2 - x). i tried letting u = pi/2 - x and i got:
    \int_0^{\pi/2} \frac{du}{1+(\cot u)^{\sqrt{2}}}

    i'm not sure what to do next though. my substitution didn't seem to do much.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}

    This problem was found on an old post and i thought it was an interesting problem. a hint was given which said remember that tan x = cot(pi/2 - x). i tried letting u = pi/2 - x and i got:
    \int_0^{\pi/2} \frac{du}{1+(\cot u)^{\sqrt{2}}}

    i'm not sure what to do next though. my substitution didn't seem to do much.
    This is famous. Write it as \int_0^{\frac{\pi}{2}}\frac{\cos^{\sqrt{2}}(x)}{\c  os^{\sqrt{2}}(x)+\sin^{\sqrt{2}}(x)}dx and let z=\frac{\pi}{2}-x and add the two integrals together.
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  3. #3
    Super Member Random Variable's Avatar
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     \int^{\frac{\pi}{2}}_{0} \frac{dx}{1+\tan^{\sqrt{2}} x} = \int^{\frac{\pi}{2}}_{0} \frac{\cos ^{\sqrt{2}}x \ dx}{\cos ^{\sqrt{2}} x+ \sin^{\sqrt{2}}x}

    let  x = \frac{\pi}{2} -u

     = - \int_{\frac{\pi}{2}}^{0} \frac{\cos ^{\sqrt{2}} (\pi/2 -u) \ du}{\cos ^{\sqrt{2}} (\pi /2 -u )+ \sin^{\sqrt{2}} (\pi /2 -u)}

     = \int^{\frac{\pi}{2}}_{0} \frac{\sin ^{\sqrt{2}} u \ du}{\sin ^{\sqrt{2}} u + \cos^{\sqrt{2}} u}

    so  \int^{\frac{\pi}{2}}_{0} \frac{dx}{1+\tan^{\sqrt{2}} x} \ dx = \frac{1}{2} \Big( \int^{\frac{\pi}{2}}_{0} \frac{\cos ^{\sqrt{2}}x \ dx}{\cos ^{\sqrt{2}}  x+ \sin^{\sqrt{2}}x}   + \int^{\frac{\pi}{2}}_{0} \frac{\sin ^{\sqrt{2}} x \ dx}{\sin ^{\sqrt{2}} x + \cos^{\sqrt{2}} x} \Big)

     = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \ dx = \frac{\pi}{4}
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  4. #4
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    Other integrals using the similar technique :

     \int^{\pi}_0 \frac{dx}{ 1 + (\sin(x))^{\cos(x)}}

    and this

     \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    Other integrals using the similar technique :

     \int^{\pi}_0 \frac{dx}{ 1 + (\sin(x))^{\cos(x)}}

    and this

     \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}
    i'm having some trouble with the first integral you posted. i tried turning sin(x) into cos(x)/cot(x) and then using the substitution x = pi/2 - u, i then added the two integrals together and divided by 2. but i couldn't simplify the integral like shown above for the first integral i posted. i also tried making some u substitutions first but all they did was make the integral more complicated. how should this integral be tackled?
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by oblixps View Post
    i'm having some trouble with the first integral you posted. i tried turning sin(x) into cos(x)/cot(x) and then using the substitution x = pi/2 - u, i then added the two integrals together and divided by 2. but i couldn't simplify the integral like shown above for the first integral i posted. i also tried making some u substitutions first but all they did was make the integral more complicated. how should this integral be tackled?
    let  x = \pi - u
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  7. #7
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    Quote Originally Posted by Random Variable View Post
    let  x = \pi - u
    wow that was so simple i can't believe i didn't think of that. thanks! now to attempt his other integral.
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  8. #8
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    would the answer to simplependulum's second integral be pi/2?
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  9. #9
    Super Member Random Variable's Avatar
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    It should be  \frac{\pi}{4} .

    EDIT: Make the substitution  x = \tan u , then the integral becomes  \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u}
    Last edited by Random Variable; May 20th 2010 at 11:03 PM.
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  10. #10
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     \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )}
    put x\mapsto\frac1x and the integral becomes \int_{0}^{\infty }{\frac{x^{\ln ^{2}x}}{\left( x^{2}+1 \right)\left( 1+x^{\ln ^{2}x} \right)}\,dx}, now sum the original integral and this one to get the result.
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  11. #11
    Super Member Random Variable's Avatar
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    or

     \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )} = \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u}  =  \int^{\frac{\pi}{2}}_{0} \frac{\cos^{\ln^{2} \tan u}u }{\cos ^{\ln^{2} \tan u} u+ \sin^{\ln^{2 \tan u} }u} \ du


    let  x = \frac{\pi}{2} - v


     = \int^{\frac{\pi}{2}}_{0} \frac{\sin^{\ln^{2} \tan v}v }{\sin^{\ln^{2}  \tan v} u+ \cos^{\ln^{2 \tan v} }v} \ dv


    since  \ln^{2} \tan v = (\ln \sin v - \ln \cos v)^{2} = (\ln \cos v - \ln \sin v)^{2} = \ln^{2} \cot v

    so it's exactly like the original problem
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  12. #12
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    Quote Originally Posted by Random Variable View Post
    or

     \int_0^{\infty} \frac{dx}{ (x^2 + 1) ( 1 + x^{(\ln(x))^2} )} = \int^{\pi/2}_{0} \frac{du}{1+ \tan^{\ln^{2} \tan u}u}  =  \int^{\frac{\pi}{2}}_{0} \frac{\cos^{\ln^{2} \tan u}u }{\cos ^{\ln^{2} \tan u} u+ \sin^{\ln^{2 \tan u} }u} \ du


    let  x = \frac{\pi}{2} - v


     = \int^{\frac{\pi}{2}}_{0} \frac{\sin^{\ln^{2} \tan v}v }{\sin^{\ln^{2}  \tan v} u+ \cos^{\ln^{2 \tan v} }v} \ dv


    since  \ln^{2} \tan v = (\ln \sin v - \ln \cos v)^{2} = (\ln \cos v - \ln \sin v)^{2} = \ln^{2} \cot v

    so it's exactly like the original problem
    oops i was careless, i made those substitutions but forgot the factor of 1/2 outside the integral. thanks!
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