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    (Original post by Benjy100)
    Problem 33

    Determine \int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx
    Solution 33

    \displaystyle \int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx= \int^a_b \frac{dx}{\sqrt{ \left( \frac{a-b}{2} \right) ^2 - \left( x-\frac{a+b}{2} \right) ^2 }} - upon expanding, completing the square and simplifying.

    Now we could use a trig. substitution, but recognise that this is just a standard integral and so

    \displaystyle = \left[ \arcsin \left( \frac{2x-a-b}{a-b} \right) \right]_b^a

    \displaystyle = \arcsin \left( \frac{a-b}{a-b} \right) - \arcsin \left( \frac{b-a}{a-b} \right)

    \displaystyle = \arcsin 1 - \arcsin (-1) = 2\arcsin 1 since arcsin is an odd function.

    \displaystyle = 2 \times \frac{\pi }{2} = \pi
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    (Original post by Lord of the Flies)
    Solution 33

    Method 1:

    Letting x=b\cos^2 t+a\sin^2 t gives:

    \displaystyle \int_b^a \frac{dx}{\sqrt{(a-x)(x-b)}} = \int_0^{\pi/2} 2 dt =\pi

    Method 2:

    \displaystyle\int_b^a \frac{1}{\sqrt{(a-x)(x-b)}}\;dx=i\int_b^a \frac{1}{\sqrt{(a-x)(b-x)}}\;dx


    =\displaystyle -2i\int_b^a\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)\;dx


    =-2i\ln(\sqrt{a-x}+\sqrt{b-x})|_b^a=-i\ln\left(\dfrac{a-b}{b-a}\right)=\pi

    (taking the principal branch - I'm sure that can be justified)
    Too quick!
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    Solution 33
    \displaystyle\int_{b}^{a} \dfrac{1}{\sqrt{(x-b)(a-x)}} dx = \displaystyle\int_{b}^{a} \dfrac{1}{ \sqrt{ \frac{1}{4} (a-b)^{2} - [x-\frac{1}{2}(a+b)]^{2}}} dx

    Letting x - \frac{1}{2} (a+b) = \frac{1}{2} (a-b) \sin \theta we get

    \displaystyle\int_{b}^{a} \dfrac{1}{ \sqrt{ \frac{1}{4} (a-b)^{2} - [x-\frac{1}{2}(a+b)]^{2}}} dx =  \displaystyle\int_{-\pi /2}^{\pi /2} d \theta  = \pi
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    (Original post by aznkid66)
    Lol, way to trip right in front of the finish line..

    EDIT: Right, ignore this then.
    You could've been a little nicer about it, even when you thought he was wrong

    Do you see why und was right though?

    Spoiler:
    Show
    \lim_{n \rightarrow \infty} (1 + \frac{a}{n})^n = e^{a}
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    (Original post by Lord of the Flies)
    ...
    (Original post by Star-girl)
    ...
    Both too quick ;_;
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    (Original post by und)
    Problem 32

    Find all positive integers n such that 12n-119 and 75n-539 are both perfect squares.
    By any chance is your next question going to be:

    Spoiler:
    Show


    A triangle has sides of length at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.

    • Thread Starter
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    (Original post by metaltron)
    By any chance is your next question going to be:

    Spoiler:
    Show


    A triangle has sides of length at most 2, 3 and 4 respectively. Determine, with proof, the maximum possible area of the triangle.

    Nah, I wasn't planning to post that one.
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    Problem 34*

    Prove that for all primes p>3, \ 24|(p^2-1).
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    Solution 34:

    p^2-1 = (p-1)(p+1)

    P is prime and greater than 3, so p-1 and p+1 must contain factors of 2, 3 and 4, so it is divisible by 24.
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    (Original post by Star-girl)
    Problem 34*

    Prove that for all primes p>3, \ 24|(p^2-1).
     p^2 - 1 = (p+1)(p-1)

    Since p>3, p cannot be divisible by 2 or 3. Hence, both p-1 and p+1 are even and one must be divisible by 4. Also, one of the three numbers, not p obviously, must be divisible by 3. Hence the whole expression is divisible by:

     2 \times 4 \times 3 = 24

    Edit : Dammit, I typed this so quickly as well!!
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    (Original post by metaltron)
     p^2 - 1 = (p+1)(p-1)

    Since p>3, p cannot be divisible by 2 or 3. Hence, both p-1 and p+1 are even and one must be divisible by 4. Also, one of the three numbers, not p obviously, must be divisible by 3. Hence the whole expression is divisible by:

     2 \times 4 \times 3 = 24

    Edit : Dammit, I typed this so quickly as well!!


    (Let me have this one, I haven't got to any of the other ones first yet! :lol:)
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    (Original post by DJMayes)


    (Let me have this one, I haven't got to any of the other ones first yet! :lol:)
    Either have I Of course you can have it, you beat me to it!
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    I was just casually typing up the solution, but you guys... :rolleyes:
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    (Original post by und)
    I was just casually typing up the solution, but you guys... :rolleyes:
    ... were not doing it casually.
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    ^^ I still haven't got to any one of them first.

    Also nicely done, DJ and metaltron.
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    Problem 35*

    Evaluate \displaystyle \int_0^{\frac{\pi}{2}} \bigg(\frac{x}{\sin x}\bigg)^2dx
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    Problem 36: */**

    A particle is projected from the top of a plane inclined at an angle  \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection  \theta from the horizontal must satisfy:

     \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}
    • Thread Starter
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    (Original post by DJMayes)
    Problem 36: */**

    A particle is projected from the top of a plane inclined at an angle  \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection  \theta must satisfy:

     \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}
    Isn't this a STEP question you were discussing recently?
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    (Original post by und)
    Isn't this a STEP question you were discussing recently?
    No; there was one involving firing up an inclined plane but it wasn't this. I did go to put a question on this thread about proving the maximum range of a projectile occurred when the angle of projection was 45 degrees but figured that it was too easy so messed about a bit and came up with this.
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    (Original post by Lord of the Flies)
    Problem 35*

    Evaluate \displaystyle \int_0^{\frac{\pi}{2}} \bigg(\frac{x}{\sin x}\bigg)^2dx
    Solution 35
    I hope!
    Solution 35
    \displaystyle\int_{0}^{\pi /2} \left( \dfrac{x}{\sin x } \right)^{2} dx

    By parts we obtain \displaystyle\int_{0}^{\pi /2} x^{2} \csc^{2} x dx = \left[-x^{2} \cot x \right]_{0}^{ \pi /2} + 2 \displaystyle\int_{0}^{\pi /2} x \cot x dx = 2 \displaystyle\int_{0}^{\pi /2} x \cot x dx

    Again, by parts, we obtain \displaystyle\int_{0}^{\pi /2} x \cot x dx = \left[x \ln (\sin x) \right]_{0}^{\pi /2} - \displaystyle\int_{0}^{\pi /2} \ln (\sin x) dx = -\displaystyle\int_{0}^{\pi /2} \ln (\sin x) dx

    \displaystyle\int_{0}^{\pi /2} x^{2} \csc^{2} x dx = -2 \times \displaystyle\int_{0}^{\pi /2} \ln (\sin x) dx

    Let \displaystyle\int_{0}^{\pi /2} \ln (\sin x) dx = I

    x \to \dfrac{\pi}{2} - x \Rightarrow \displaystyle\int_{0}^{\pi /2} \ln (\sin x) dx = \displaystyle\int_{0}^{\pi /2} \ln (\cos x) dx

    2I = \displaystyle\int_{0}^{\pi /2} \ln (\sin 2x) - \ln 2 dx

    Let t = 2x \Rightarrow \displaystyle\int_{0}^{\pi /2} \ln (\sin 2x) dx = \dfrac{1}{2} \displaystyle\int_{0}^{\pi} \ln (\sin t) dt

    \displaystyle\int_{0}^{\pi} \ln (\sin t) dt = 2 \displaystyle\int_{0}^{\pi /2} \ln (\sin t) dt \Rightarrow \dfrac{1}{2} \displaystyle\int_{0}^{\pi} \ln (\sin t) dt = \displaystlye\int_{0}^{\pi /2} \ln (\sin t) dt = I

    2I = I - \displaystyle\int_{0}^{\pi /2} \ln 2 dx \Rightarrow I = - \dfrac{\pi}{2} \ln 2

    \Rightarrow \displaystyle\int_{0}^{\pi /2} \left( \dfrac{x}{\sin x } \right)^{2} dx = -2 \times -\dfrac{\pi}{2} \ln 2 = \pi \ln 2
 
 
 
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