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Math Help - Integration

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Integration

    Could someone post on here a semi-difficult integral if they wouldn't mind
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  2. #2
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Could someone post on here a semi-difficult integral if they wouldn't mind
    \int_\frac{-\pi}{4}^\frac{\pi}{4} \arccos(\tan x)\;dx

    i don't know if that's even remotely difficult for you. it was semi-difficult for me.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    \int \arccos(\tan x)\;dx

    i don't know if that's even remotely difficult for you. it was semi-difficult for me.
    This integral does not have a elementary solution? You know that right? But I greatly extremely appreciate your effort!
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    This integral does not have a elementary solution? You know that right? But I greatly extremely appreciate your effort!
    try it out....i forgot the intervals
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  5. #5
    Super Member PaulRS's Avatar
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    Calculate \int_0^{\infty}{\frac{\sin^2(x)}{x^2}dx} bearing in mind that \int_0^{\infty}{\frac{\sin(x)}{x}dx}=\frac{\pi}{2}

    It's not difficult but I still like the result
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by polymerase View Post
    try it out....i forgot the intervals
    Oh...that makes a big difference

    Well lets see here \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcos(tan(x))dx=\frac{\pi{  x}}{2}\bigg|_{\frac{-\pi}{4}}^{\frac{\pi}{4}}-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}arcsin(tan(x))dx

    Simplifying and adding the integrals we get an indentity \frac{{\pi}^2}{4}=\frac{{\pi}^2}{4}

    so the answer is \frac{{\pi}^2}{4}

    How did you do it? Modifying the phase shift?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Calculate \int_0^{\infty}{\frac{\sin^2(x)}{x^2}dx} bearing in mind that \int_0^{\infty}{\frac{\sin(x)}{x}dx}=\frac{\pi}{2}

    It's not difficult but I still like the result
    Hmm...I know the way you are thinking of but I will try this

    \int_0^{\infty}\frac{\sin^2(x)}{x^2}dx=\frac{1}{2}  \int_0^{\infty}\frac{1-\cos(2x)}{x^2}dx

    seperating we get \frac{1}{2}\bigg[\int_0^{\infty}\frac{1}{x^2}dx-\int_0^{\infty}\frac{\cos(2x)}{x^2}dx\bigg]

    Now I dont feel like writing it all but I used power series and got \frac{\pi}{2} as the answer as well
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Are you sure?

    About the integral I posted: You should make use of some new techniques you've seen in this forum not long ago
    Two things...what is wrong about that...that is the answer isnt it? I even checked my calculator once I was done ...and please dont tell me I have to use the imaginary part of e^{-x(a-bi)} I am only seventeen and a junior in highschool...dont give me this complex analysis stuff lol
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  9. #9
    Super Member PaulRS's Avatar
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    Define J(p)=\int_{0}^{\infty}{\frac{\sin^2(p\cdot{x})}{x^  2}dx} ...

    ...and now differentiate under the integral sign
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Define J(p)=\int_{0}^{\infty}{\frac{\sin^2(p\cdot{x})}{x^  2}dx} ...

    ...and now differentiate under the integral sign
    Haha what part about junior in highschool, seventeen, does not know complex analysis...uhm...I will try....lets see here we want...Give me a little more hint...is this something to do with Im[e^{ix}]?
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  11. #11
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Haha what part about junior in highschool, seventeen, does not know complex analysis...uhm...I will try....lets see here we want...Give me a little more hint...is this something to do with Im[e^{ix}]?
    Don't worry

    Leibniz's integral rule


    Differentiation under the integral sign - Wikipedia, the free encyclopedia

    You do not need to use complex numbers
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Ok so if J(p)=\int{\frac{\sin^2(px)}{x^2}}dx

    then J'(p)=2\int{\frac{\sin(ax)\cos(ax)}{x}}dx?
    That is what I am getting from that...I took the partial in respect to a?
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  13. #13
    Super Member PaulRS's Avatar
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    Here is an example of Leibniz's Rule in use

    http://www.mathhelpforum.com/math-he...77-post19.html

    And here is the complete thread: http://www.mathhelpforum.com/math-he...tegrals-2.html
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Here is an example of Leibniz's Rule in use

    http://www.mathhelpforum.com/math-he...77-post19.html

    And here is the complete thread: http://www.mathhelpforum.com/math-he...tegrals-2.html
    So is what I did completely wrong?
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  15. #15
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok so if J(p)=\int{\frac{\sin^2(px)}{x^2}}dx

    then J'(p)=2\int{\frac{\sin(ax)\cos(ax)}{x}}dx?
    That is what I am getting from that...I took the partial in respect to a?
    It should have been: J'(p)=\int_0^{\infty}{\frac{2\sin(px)\cos(px)}{x}}  dx

    Now relate it to \int_0^{\infty}{\frac{\sin(x)}{x}}dx

    And finally integrate: \int_0^tJ'(p)dp=J(t)-J(0)=J(t)
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