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    (Original post by k9markiii)
    I liked that middle question can't be bothered with the last one.
    For question 362, I made an edit.

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    (Original post by james22)
    That's the one I meant, I'm curious what the solution is because I could not find any way to make simpler. I managed to convert it to several different integrals but they didn't look any nicer.
    Well that's pretty much what I did, though it seemed satisfying so I'll type it up anyway:

    (Original post by Lord of the Flies)
    Problem 260***

    \displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx
    Let \displaystyle f(\alpha)= \int_0^{\infty} \frac{\sin( \alpha \tan x)}{x} \ dx so that \displaystyle \int_{-\infty}^{\infty} x^{-1} \tan x \cos \tan x \ dx = 2 f'(1)

    Taking the Laplace transform, applying Fubini's theorem and then using the standard result (proven by IBP or the exponential representation of sine) for the Laplace transform of \sin ax, we have

    \displaystyle \mathcal{L}\{f(\alpha)\} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-s \alpha} \sin( \alpha \tan x) \ d \alpha \ dx = \int_0^{\infty} \frac{1}{x} \left( \frac{\tan x}{s^2+\tan^2 x} \right) \ dx

    Let \displaystyle g(\beta) = \int_0^{\infty} \frac{1}{x} \left( \frac{\tan \beta x}{s^2+\tan^2 x} \right) \ dx so that \displaystyle g(1) = \mathcal{L}\{f(\alpha)\}

    \displaystyle \Rightarrow g'(\beta) = \int_0^{\infty} \frac{\sec^2 \beta x}{s^2+\tan^2 x} \ dx = \int_0^{\infty} \frac{\sec^2 \beta x}{(s^2-1)+\sec^2 x} \ dx

    The starter question on an old USAMO paper:

    Problem 366*

    Find the average of the numbers \displaystyle n \sin n^{\circ} (n=2, 4, 6, \cdots, 180)
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    I have missed this integral.

    Solution 260

    Let me use the notation of Jkn's post.
    \begin{aligned} \displaystyle \int_{0}^{\pi(n+\frac{1}{2})} \frac{\sin( \alpha \tan x)}{x}dx &= \frac{1}{2}\sum_{k=-n}^{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin( \alpha \tan x)}{x+k\pi}dx \\&= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sum_{k=-n}^{n} \frac{x}{x^{2}-k^{2}\pi^{2}} \right)\sin( \alpha \tan x)dx \end{aligned}.
    Note that \displaystyle \sum_{k=-n}^{n} \frac{x}{x^{2}-k^{2}\pi^{2}} = \cot x + O(\frac{1}{n}) .
    Hence, we can let n \to \infty and obtain
    \begin{aligned} \displaystyle f(\alpha)& = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot x \sin(\alpha \tan x)dx \\& \mathop= \limits^{\tan x \mapsto x} \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin( \alpha x)}{x(1+x^{2})} dx \\&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(\alpha x)}{x} dx -\frac{1}{2}\int_{-\infty}^{\infty} \frac{x\sin(\alpha x)}{1+x^{2}}dx  = \frac{\pi}{2} - \frac{\pi}{2}e^{-\alpha} \end{aligned}.
    The last integral is an easy exercise in complex analysis.
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    (Original post by Mladenov)
    \displaystyle \int_{0}^{\pi(n+\frac{1}{2})} \frac{\sin( \alpha \tan x)}{x}dx &= \frac{1}{2}\sum_{k=-n}^{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin \alpha \tan x}{x}dx
    What's this supposed to be? (its's a typo!)
    \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin( \alpha x)}{x(1+x^{2})} dx = \int_{-\infty}^{\infty} \frac{\sin(\alpha x)}{x} dx -\int_{-\infty}^{\infty} \frac{x\sin(\alpha x)}{1+x^{2}}dx  = \frac{\pi}{2} - \frac{\pi}{2}e^{-\alpha}
    You missed out the \frac{1}{2}s. Typo?
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    (Original post by Arieisit)
    Problem 362*

    This is actually very similar to an A level question but without the added guidance that they usually provide.

    A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter x represents the length of the rectangular section and r represents the radius of the semicircle.
    Determine the length, x, that maximises the area enclosed by the track.

    EDIT

    The perimeter of the track must be 600 metres.

    I completely forgot to put that in. Sorry

    600m running track hmm OK different from the 400m standard, nice. Well A=\pi r^2 + 2rx and P = 2x+2\pi r where P is 600m. There are a number of ways to do this but it would be neatest with lagrange multipliers I think.

    A=\pi r^2 + 2rx

    0 = 2x+2\pi r - 600

    A = \pi r^2 + 2rx + \lambda(2x+2\pi r - 600)

    A = \pi r^2 + 2rx + 2\lambda x +2\lambda \pi r - 600\lambda

    \frac{dA}{dx} = \frac{dA}{dr} = \frac{dA}{d\lambda} = 0

    0 = 2 x +2 \pi r - 600

    0 = 2\pi r+ 2x +2\lambda \pi

    0 = 2r + 2\lambda

    r = - \lambda

    0 = 2\pi r +2x -2\pi r

    0 = 2x

    This implies the maximum area is enclosed when x = 0.

    This makes sense as a circle contains the most area for a given perimeter of any shape.
    Certainly didn't go for a straightforward method but using langrange multipliers is useful when you are adding constraints to a system. It seems an odd answer are you sure you've got the question right?
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    (Original post by Jkn)
    What's this supposed to be? (its's a typo!)

    You missed out the \frac{1}{2}s. Typo?
    Indeed. What I did is \displaystyle \int_{n\pi - \frac{\pi}{2}}^{n\pi+\frac{\pi}{  2}} \frac{\sin(\alpha \tan x)}{x}dx = \int_{ -\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x+n\pi}dx.
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    (Original post by Mladenov)
    Indeed. What I did is \displaystyle \int_{n\pi - \frac{\pi}{2}}^{n\pi+\frac{\pi}{  2}} \frac{\sin(\alpha \tan x)}{x}dx = \int_{ -\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x+n\pi}dx.
    Wow, that's brilliant! Beautiful solution!
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    Sorry to be an anus but the first expression is still typed wrong! And the half shouldn't come in to the second line (after adding it with the modified expression)


    Btw, I'm currently fiddling around trying to see if I can find a closed form for \displaystyle \mathcal{L}\{ \log^n (x) \} and have come to a halt at \displaystyle \frac{d^n}{dx^n} \Gamma(x+1) \Big|_{x=0} which, despite being able to evaluate \phi^{(n)}(1) very neatly, I am yet to find a general expression for it (it decays into a mess of polygammas). I've also realised that, if this could be evaluated, we would have a taylor series for the gamma function. Is there a taylor series for the gamma function or am I going into the realm of the unknown?
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    (Original post by k9markiii)
    Certainly didn't go for a straightforward method but using langrange multipliers is useful when you are adding constraints to a system. It seems an odd answer are you sure you've got the question right?
    Yes, I am sure this time. My solution is much simpler but I agree, the answer is definitely odd.
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    (Original post by Arieisit)
    Yes, I am sure this time. My solution is much simpler but I agree, the answer is definitely odd.
    I was practising out some fun maths. I know you can do it through simultaneous equations but I was practising some maths I learnt last year to get ready for lectures starting next week.
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    (Original post by Arieisit)
    Problem 362*

    This is actually very similar to an A level question but without the added guidance that they usually provide.

    A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter x represents the length of the rectangular section and r represents the radius of the semicircle.
    Determine the length, x, that maximises the area enclosed by the track.

    EDIT

    The perimeter of the track must be 600 metres.

    I completely forgot to put that in. Sorry
    You don't need to assign a numerical P here, so long as P is constant.

    I suppose you have marked this as * so maybe you just want to show something nice - but surely the answer is trivial when you consider what a circle actually is and what some of its inherent properties are.
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    (Original post by Llewellyn)
    You don't need to assign a numerical P here, so long as P is constant.

    I suppose you have marked this as * so maybe you just want to show something nice - but surely the answer is trivial when you consider what a circle actually is and what some of its inherent properties are.
    I think you quoted the wrong person. Lol

    EDIT

    I reread your post. I was looking for a numerical answer hence P was given.

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    (Original post by Arieisit)
    Problem 362*

    This is actually very similar to an A level question but without the added guidance that they usually provide.

    A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter x represents the length of the rectangular section and r represents the radius of the semicircle.
    Determine the length, x, that maximises the area enclosed by the track.

    EDIT

    The perimeter of the track must be 600 metres.
    Ignore this: I misread the question and maximised the area for the rectangle surmounted by a semi circle on either side
    .
    Wrong solution
    Solution 362

    Let A be the area of the track and let P be its perimeter. That means that:

    A = \pi r^2 + 2rx

    P = 600 = 2x + 2\pi r

    To find the length of x that maximises the area enclosed by the track, it'd be helpful to find the area in terms of x alone.

    A = \dfrac{1}{\pi} \left( 9000 - x^2 \right)

    \dfrac{dA}{dx} = \dfrac{-2x}{\pi}

    The maximum value of the area occurs when the first derivative is equal to 0 and the second derivative is less than 0.

    \dfrac{dA}{dx} = 0 \Rightarrow x=0

    \dfrac{d^2A}{dx^2} = \dfrac{-2}{\pi} < 0

    \therefore The area is maximised when the length x is equal to 0 and the track is circular.
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    (Original post by Khallil)
    Solution 362

    Let A be the area of the track and let P be its perimeter. That means that:

    A = \pi r^2 + 2rx

    P = 600 = 2x + 2\pi r

    To find the length of x that maximises the area enclosed by the track, it'd be helpful to find the area in terms of x alone.

    A = \dfrac{1}{\pi} \left( 9000 - x^2 \right)

    \dfrac{dA}{dx} = \dfrac{-2x}{\pi}

    The maximum value of the area occurs when the first derivative is equal to 0 and the second derivative is less than 0.

    \dfrac{dA}{dx} = 0 \Rightarrow x=0

    \dfrac{d^2A}{dx^2} = \dfrac{-2}{\pi} < 0

    \therefore The area is maximised when the length x is equal to 0 and the track is circular.
    I have a slight qualm - your equations are for a circle when it is quite specifically a semi-circle that is mentioned.
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    (Original post by Arieisit)
    Problem 362*
    (Original post by DJMayes)
    I have a slight qualm - your equations are for a circle when it is quite specifically a semi-circle that is mentioned.
    Damn it! I read the question but my brain told me that the rectangle is surmounted by a semi circle on either side, like the race tracks in the Olympics

    I hope this is right:

    Solution 362

    Let A be the area of the track and let P be its perimeter. That means that:

    A = \dfrac{1}{2} \pi r^2 + 2rx

    P = 600 = 2r + 2x + \pi r

    To find the length of x that maximises the area enclosed by the track, it'd be helpful to find the area in terms of x alone.

    A = \dfrac{1}{(2+\pi)^2} \left( 180000\pi + 2400x - x^2 \left(8+2\pi \right) \right)

    \dfrac{dA}{dx} = \dfrac{2400- 2x(8+2\pi)}{(2+\pi)^2}

    The maximum value of the area occurs when the first derivative is equal to 0 and the second derivative is less than 0.

    \dfrac{dA}{dx} = 0 \ \Rightarrow \ 2400=2x(8+2\pi) \ \Rightarrow \ 1200=x(8+2\pi) \ \Rightarrow \ x = \dfrac{600}{4+\pi}

    \dfrac{d^2A}{dx^2} = \dfrac{-2(8+2\pi)}{(2+\pi)^2} < 0

    \therefore The area is maximised when the length x is equal to \dfrac{600}{4+\pi}
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    Since nobody attempted Problem 364 I'll provide a hint or two.

    Problem 364***

    Determine the extreme values of

    f(x,y,z,u,v,w) = \frac{1}{1+x+u} + \frac{1}{1+y+v} + \frac{1}{1+z+w}


    where xyz = a^3, uvw = b^3, and x,y,z,u,v,w>0.

    Spoiler:
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    Find all the intrinsic critical points of f on M
    Calculate \begin{array}{l} sup \\ M \end{array} f.
    Investigate whether or not f attains the value\begin{array}{l} sup \\ M \end{array} f at some point of M



    Problem 367**

    If I_n = \displaystyle\int^\frac{\pi}{4}_  0 tan^nx\ dx,  n \geq 2 find I_4

    Problem 368**

    Given that I_m = \displaystyle\int (cos^mx) (sin3x) dx and J_m = \displaystyle\int (cos^mx) (sin2x) dx, prove that (m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x).
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    (Original post by Arieisit)

    Problem 368**

    Given that I_m = \displaystyle\int (cos^mx) (sin3x) dx and J_m = \displaystyle\int (cos^mx) (sin2x) dx, prove that (m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x).
    \displaystyle I_{m}=\int \underbrace{ \cos^m x }_{\mathrm{u}} \underbrace{ \sin 3x }_{\mathrm{v'}} \,dx

    \displaystyle I_{m}=- \frac{1}{3} \cos 3x \cos^m x -\frac{m}{3} \int (\sin x \cos 3x) \cos^{m-1} x \, dx

    But note \sin (3x-x) = \sin 3x \cos x - \cos 3x \sin x

    So \cos 3x \sin x = \sin 3x \cos x - \sin 2x

    So \displaystyle 3I_{m}=-\cos 3x \cos^m x -m \left( I_m - J_{m-1} \right)

    So (m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x)
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    Was my response to problem 362 correct?

    (Original post by Arieisit)
    Problem 367**

    If I_n = \displaystyle\int^\frac{\pi}{4}_  0 tan^nx\ dx,  n \geq 2 find I_4
    Solution 367

    \begin{aligned} I_n & = \displaystyle\int^\frac{\pi}{4}_  0 tan^nx\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_  0 tan^2x \ tan^{n-2}x\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_  0 \left( sec^2x - 1 \right) tan^{n-2}x\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_  0 sec^2x \ tan^{n-2}x\ dx - I_{n-2} \end{aligned}

    u=tan x \ \Rightarrow \ \dfrac{du}{dx} = sec^2x

    x=0 \Rightarrow u=0, \ x=\dfrac{\pi}{4} \Rightarrow u=1

    \begin{aligned} \therefore I_n & = \displaystyle\int^{1}_0 u^{n-2}\ du - I_{n-2} \\ & = \Big[ \dfrac{u^{n-1}}{n-1} \Big]_{0}^1 - I_{n-2} = \dfrac{1}{n-1} - I_{n-2} \end{aligned}

    \begin{aligned} I_4 & = \frac{1}{3} - I_2 = \frac{1}{3} - \left( 1 - I_0 \right) = I_0 - \frac{2}{3} = \frac{\pi}{4} - \frac{2}{3} \end{aligned}
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    [QUOTE=Khallil;44692928]Was my response to problem 362 correct?

    I thought this thread wasn't about getting the right answer

    I got a similar answer if that helps :rolleyes:

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    (Original post by Arieisit)
    (Original post by Khallil)
    Was my response to problem 362 correct?
    I thought this thread wasn't about getting the right answer
    Indeed it isn't

    (Original post by Arieisit)
    I got a similar answer if that helps :rolleyes:
    How similar? :pierre:
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    (Original post by Khallil)
    How similar? :pierre:
    Lets just say that yours looks more elegant.



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