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    Just read about the MIT integration bee! Turns out the integrals on there aren't as hard as the ones on here... :lol:

    The qualifying round is basically having 20 minutes to solve as many of a list of 25 integrals as possible and then, in the finals, you do the same but neck and neck with someone else of chalk boards side by side. Most of them seem mad easy :lol:

    Here's a link if anyone's interested... or want to post more on here!


    I come bearing gifts

    Problem 174*

    Evaluate \displaystyle \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx

    Problem 175*

    Evaluate \displaystyle\int_0^1 \sin(\arccos(x)) dx

    Problem 176*

    Evaluate \displaystyle\int_0^1 x(1-x)^{99} dx

    Problem 177*

    Evaluate \displaystyle\int \sqrt{\csc(x)-\sin(x)} dx

    Problem 178*

    Evaluate \displaystyle\int_0^2 x^5 \sqrt{1+x^3} dx

    Problem 179*

    Evaluate \displaystyle \int_0^{\frac{\pi}{2}} \frac{\sin(4x)}{\sin(x)} dx

    Problem 180*

    Evaluate \displaystyle \int \left( \cos(x) \ln(x) +\frac{\sin(x)}{x} \right) dx

    Problem 181*

    Evaluate \displaystyle \int \frac{1}{\sqrt{x}-1} dx

    Problem 182*

    Evaluate \displaystyle \int \frac{1}{(1+\sqrt{x})\sqrt{x-x^2}} dx

    Problem 183*

    Evaluate \displaystyle \int \frac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx

    ...and finally, this one was tie breaker between the top two (how it took them a whole minute I have no idea! You can do this **** in your head :lol:)

    Problem 184*

    Evaluate \displaystyle \int \frac{x^4}{1-x^2} dx
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    Solution 174

    Let x = \tan \theta :

    \displaystyle\int_{0}^{\infty} \dfrac{\ln x}{1 + x^{2}} dx = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \tan \theta \right) d \theta

    = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \sin \theta \right ) d \theta - \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \cos \theta \right) d \theta = 0

    Solution 175

    \displaystyle\int_{0}^{1} \sin \left( \arccos x \right) dx = \displaystyle\int_{0}^{1} \sqrt{1 - x^{2}} dx

    Let x = \sin \theta :

    \displaystyle\int_{0}^{1} \sqrt{1 - x^{2}} dx = \dfrac{1}{2} \displaystyle\int_{0}^{ \frac{\pi}{2}} 1 + \cos 2 \theta \ d \theta = \dfrac{\pi}{4}

    Solution 177

    \displaystyle\int \sqrt{ \csc x - \sin x} dx = \displaystyle\int \dfrac{ \cos x}{\sqrt{ \sin x}} dx = 2 \sqrt{ \sin x } + \mathcal{C}

    Solution 180

    \displaystyle\int \left( \ln x \cdot  \cos x + \dfrac{ \sin x}{x} \right) dx = \ln x \cdot \sin x  + \mathcal{C}

    I'll leave the rest, 183 looks like an algebraic mess :lol:
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    Solution 175:

    I =  \int_0^1 sin(arccosx) \ dx

    Let  arcsinu = arccosx

     \dfrac{1}{\sqrt{1-u^2}} du = -\dfrac{1}{\sqrt{1-x^2}} dx

    Re-arranging, we get:

     dx = -\dfrac{\sqrt{1-x^2}}{\sqrt{1-u^2}} du

     \sqrt{1-x^2} = \sqrt{1-cos^2arcsinu} = sinarcsinu = u

     \Rightarrow I = - \int_1^0 \dfrac{u^2}{\sqrt{1-u^2}} \ du

     I = \int_0^1 \dfrac{ 1-\sqrt{1-u^2}^2}{\sqrt{1-u^2}} \ du

     = \int_0^1 \dfrac{1}{\sqrt{1-u^2}}- \sqrt{1-u^2} \ du

    The first part of this is a standard arcsin integral and yields  \frac{\pi }{2} . Let's look at the second part:

     J = \int_0^1 \sqrt{1-u^2} \ du

    Let  u = sinx

     \Rightarrow J = \int_0^{\frac{\pi }{2}} cos^2x \ dx

     = \frac{1}{2} \int_0^{\frac{\pi }{2}} cos2x+1 \ dx

     = \frac{1}{2} [ \frac{1}{2}sin2x+x ]_0^{\frac{\pi }{2}}

     = \frac{\pi }{4}

    Therefore our final answer is \frac{\pi }{4}
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    (Original post by Felix Felicis)
    Problem 171, 172
    Felix you need to relable your questions, there already is a 171



    Problem 185*

    Find all f:\;\mathbb{R}\to\mathbb{R} such that:

    \begin{aligned}(\text{i})\;f(x)>  0\qquad (\text{ii})\; f(x)\geq x+1\qquad(\text{iii})\; f(x)f(-x)=1\qquad (\text{iv})\;f(2x)=f^2(x)

    Problem 186*

    A nice problem. Prove/disprove that there exists f:\;\mathbb{N}\to\mathbb{N} such that:

    f^{f(n)}(n)=n+1

    f^k means f applied k times in this case.



    Well, to those who were unhappy about the lack of (*)'s, you now have loads. Go go go go!
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    (Original post by Jkn)

    Problem 178*

    Evaluate \displaystyle\int_0^2 x^5 \sqrt{1-x^3} dx
    Are you sure the upper limit is supposed to be 2? Otherwise the integral is not defined for real numbers past 1...
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    (Original post by Felix Felicis)
    Solution 174

    Let x = \tan \theta :

    \displaystyle\int_{0}^{\infty} \dfrac{\ln x}{1 + x^{2}} dx = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \tan \theta \right) d \theta

    = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \sin \theta \right ) d \theta - \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \cos \theta \right) d \theta = 0
    In light of it being a "speed" competition, allow me to direct you toward my solution :rolleyes:

    Solution 174
    (2)

    Let x \to x^{-1},

    \displaystyle \Rightarrow \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx = \int_0^{\infty} -\frac{\ln(x)}{1+x^2} dx =0 \ \square
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    (Original post by DJMayes)
    Are you sure the upper limit is supposed to be 2? Otherwise the integral is not defined for real numbers past 1...
    Sorry typo, 'tis now fixed (that's what you get for setting two questions at once) :lol:
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    (Original post by Jkn)
    In light of it being a "speed" competition, allow me to direct you toward my solution :rolleyes:

    Solution 174
    (2)

    Let x \to x^{-1},

    \displaystyle \Rightarrow \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx = \int_0^{\infty} -\frac{\ln(x)}{1+x^2} dx =0 \ \square
    It takes like 10 seconds less :dontknow: Neat, nonetheless :lol:
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    (Original post by Lord of the Flies)
    Felix you need to relable your questions, there already is a 171



    Problem 185*

    Find all f:\;\mathbb{R}\to\mathbb{R} such that:

    \begin{aligned}(\text{i})\;f(x)>  0\qquad (\text{ii})\; f(x)\geq x+1\qquad(\text{iii})\; f(x)f(-x)=1\qquad (\text{iv})\;f(2x)=f^2(x)

    Problem 186*

    A nice problem. Prove/disprove that there exists f:\;\mathbb{N}\to\mathbb{N} such that:

    f^{f(n)}(n)=n+1

    f^k means f applied k times in this case.



    Well, to those who were unhappy about the lack of (*)'s, you now have loads. Go go go go!
    No way these look exactly like the functional equations questions you get in Core 4!
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    Lol jk functional equations are not * questions :pierre:
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    (Original post by Felix Felicis)
    It takes like 10 seconds less :dontknow: Neat, nonetheless :lol:
    Meh :dontknow:

    Please post some interesting series and/or integration questions
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    (Original post by Jkn)
    Meh :dontknow:

    Please post some interesting series and/or integration questions
    I'll have a look around and see what I have :lol:

    Edit: I thought the first integral you posted was \displaystyle\int_{0}^{\infty} \dfrac{ \ln x}{(x^{2} + 1)^{2}} \ dx and I thought no way in hell can anyone do that in 20 minutes :lol:
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    (Original post by Jkn)
    Lol jk functional equations are not * questions
    Yes they are. In fact a lot of functional equations only require GCSE knowledge.
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    (Original post by Lord of the Flies)
    Yes they are. In fact a lot of functional equations only require GCSE knowledge.
    But the concept of a function isn't introduced until a-level :rollseyes:
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    (Original post by Jkn)
    But the concept of a function isn't introduced until a-level :rollseyes:
    New rule. In your head always add another * to LOTF's questions
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    (Original post by Jkn)
    But the concept of a function isn't introduced until a-level :rollseyes:
    I'm pretty sure I've seen function-related GCSE questions on the maths forum but whatever. Anyhow, can we stop cluttering the thread with this chatter, both questions are * the end.
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    (Original post by Jkn)

    Problem 178*

    Evaluate \displaystyle\int_0^2 x^5 \sqrt{1+x^3} dx
    Let  x^3 = sinh^2u

     3x^2 \ dx = 2sinhucoshu \ du

     \Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

     = \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

     = \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

     = = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

    x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

    Finally plugging those in I get 1192/45, although this isn't really a nice answer...
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    (Original post by Lord of the Flies)
    I'm pretty sure I've seen function-related GCSE questions on the maths forum but whatever.
    Yes, but the way GCSEs are taught means they wouldn't have a clue what a function was even if they do stuff related to them....
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    (Original post by DJMayes)
    Let  x^3 = sinh^2u

     3x^2 \ dx = 2sinhucoshu \ du

     \Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

     = \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

     = \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

     = = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

    x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

    Finally plugging those in I get 1192/45, although this isn't really a nice answer...
    Or u =1+x^3 ?
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    (Original post by Felix Felicis)
    Edit: I thought the first integral you posted was \displaystyle\int_{0}^{\infty} \dfrac{ \ln x}{(x^{2} + 1)^{2}} \ dx and I thought no way in hell can anyone do that in 20 minutes :lol:
    Lightweight, you can get it form applying integration by parts to the other one by integrating \ln(x) (and differentiating the denominator) and then evaluating a fairly trivial other part with a tanx substitution
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    (Original post by bananarama2)
    Or u =1+x^3 ?
    Hyperbolics are more fun.

    (What answer do you get with that sub though?)
 
 
 
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