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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by aymanzayedmannan)
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    I've just subbed \sec x itself: \displaystyle \text{let} \ u = \sec x \Rightarrow \frac{\mathrm{d} u}{\mathrm{d} x} = \sec x \tan x

    Rewriting the integral in terms of u gives us

    \displaystyle \int \frac{\sec x}{\sec x \tan x}\mathrm{d}u = \int \frac{\mathrm{d}u}{\tan x} =\int \frac{\mathrm{d}u}{\sqrt{\sec^{2  }x -1}}=\int \frac{\mathrm{d}u}{\sqrt{u^{2}-1}}

    Note that this is the standard integral for arcosh(u) so I'm quoting it without proof:

    \displaystyle \int \frac{\mathrm{d}u}{\sqrt{u^{2}-1}}= \cosh^{-1}\left ( u \right ) + \mathrm{C} = \ln\left \{ u + \sqrt{u^{2}-1} \right \} + \ln \mathrm{k} = \ln \left (\sec x + \sqrt{\sec^{2}x -1} \right ) + \ln \mathrm{k}

    by using the logarithmic form of arcosh (again, without proof)

    \displaystyle \therefore \boxed{\int \sec x \ \mathrm{d}x = \ln \left | \mathrm{k}\left ( \sec x + \tan x \right ) \right |}
    It's a shame no actual C4 students attempted this
    What's that inverse cosh?
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    (Original post by aymanzayedmannan)
    It's a shame no actual C4 students attempted this
    That's hilarious. :rofl:
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    (Original post by Dohaeris)
    What's that inverse cosh?
    Don't worry about that - it's just something you learn in FP3, you won't need it in C4! Kvothe's method is the perfect way to tackle this using C4 knowledge, and it's actually not that daunting - it's only a bit of trig manipulation and partial fractions, nothing else to it! You were probably just intimidated by the lack of structure but I assure you that these questions will have more structure in the exam and they'd most likely tell you what method to use.

    If you do want to know about the inverse cosh thing, you can look them up here. To summarise, y = \cosh x = \frac{e^{x}+e^{-x}}{2} \Longleftrightarrow \cosh^{-1}(x) = \ln\left \{ x + \sqrt{x^{2}-1} \right \}, the latter of which you can obtain by some algebraic manipulation.
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    (Original post by Student403)
    Wow that's a really good idea
    Not as cool as the one line of working Zacken did..
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    (Original post by aymanzayedmannan)
    Not as cool as the one line of working Zacken did..
    Yours is the kind of solution that makes me think

    "LOL question got rekt"
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    (Original post by aymanzayedmannan)
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    I've just subbed \sec x itself: \displaystyle \text{let} \ u = \sec x \Rightarrow \frac{\mathrm{d} u}{\mathrm{d} x} = \sec x \tan x

    Rewriting the integral in terms of u gives us

    \displaystyle \int \frac{\sec x}{\sec x \tan x}\mathrm{d}u = \int \frac{\mathrm{d}u}{\tan x} =\int \frac{\mathrm{d}u}{\sqrt{\sec^{2  }x -1}}=\int \frac{\mathrm{d}u}{\sqrt{u^{2}-1}}

    Note that this is the standard integral for arcosh(u) so I'm quoting it without proof:

    \displaystyle \int \frac{\mathrm{d}u}{\sqrt{u^{2}-1}}= \cosh^{-1}\left ( u \right ) + \mathrm{C} = \ln\left \{ u + \sqrt{u^{2}-1} \right \} + \ln \mathrm{k} = \ln \left (\sec x + \sqrt{\sec^{2}x -1} \right ) + \ln \mathrm{k}

    by using the logarithmic form of arcosh (again, without proof)

    \displaystyle \therefore \boxed{\int \sec x \ \mathrm{d}x = \ln \left | \mathrm{k}\left ( \sec x + \tan x \right ) \right |}
    It's a shame no actual C4 students attempted this
    Perhaps they did. I have attempted many Indeterminate problems before giving up in frustration. It might appear to be on the same level.

    I like your solution. I've not seen it before.
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    (Original post by Dohaeris)
    What's that inverse cosh?
    Nothing more than C3 logarithms, we define \cosh \theta = \frac{1}{2} \left(e^{\theta} + e^{-\theta}\left) so a bit of (after domain restrictions for injectivity puporses) quadratic shizzle fo shizzle, we have \text{arcosh} (\theta) = \ln (\theta + \sqrt{\theta^2 -1}), for  \theta \geq 1.
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    (Original post by Kvothe the arcane)
    Perhaps they did. I have attempted many Indeterminate problems before giving up in frustration. It might appear to be on the same level.

    I like your solution. I've not seen it before.
    Can relate lol This would look pretty difficult for a C4 student who's just starting integration.

    (Original post by Zacken)
    Nothing more than C3 logarithms, we define \cosh \theta = \frac{1}{2} \left(e^{\theta} + e^{-\theta}\left) so a bit of (after domain restrictions for injectivity puporses) quadratic shizzle fo shizzle, we have \text{arcosh} (\theta) = \ln (\theta + \sqrt{\theta^2 -1}), for  \theta \geq 1.
    I like how you used "domain restrictions for injectivity purposes" and "shizzle of shizzle" in the same sentence. :laugh:

    (Original post by Student403)
    Yours is the kind of solution that makes me think

    "LOL question got rekt"
    Take that! \int:whip2:
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    (Original post by aymanzayedmannan)
    Can relate lol This would look pretty difficult for a C4 student who's just starting integration.



    I like how you used "domain restrictions for injectivity purposes" and "shizzle of shizzle" in the same sentence. :laugh:



    Take that! \int:whip2:
    Perfect :rofl:
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    (Original post by aymanzayedmannan)


    Take that! \int:whip2:
    This might be the best thing I've seen on TSR, ever. :laugh:
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    (Original post by aymanzayedmannan)
    This standard result is quoted in the formula booklet but would anyone like to have a go at it?

    \displaystyle \int \sec x\ \mathrm{d}x
    Lol they wont ask that but try differentiating ln|sec+tan| + C and its integrand will appear apparent
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    Finally started hitting 87, 88% averages in C4 in my 1st+2nd paper now.

    Still plenty of time to grow just need more practice with drawing diagrams for vectors
    and so many past papers to do left to do.

    I honestly don't know where I would be if I didn't come across Madasmaths website = key to success :heart:

    Just looking forward to the summer so I can finish self teaching FP1-2 + more mechanics.
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    Challenging derivatives
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    (Original post by EricPiphany)
    Challenging derivatives
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    The question seem gibberish but one way perhaps..

    

ln(y-x) = -xy

\dfrac{\dfrac{dy}{dx} - 1}{y-x} = -x\dfrac{dy}{dx} -y

etc....
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    (Original post by GeologyMaths)
    The question seem gibberish but one way perhaps..

    

ln(y-x) = -xy

\dfrac{\dfrac{dy}{dx} - 1}{y-x} = -x\dfrac{dy}{dx} -y

etc....
    Looks good
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    (Original post by EricPiphany)
    Challenging derivatives
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    \ln (y-x) = -xy \Rightarrow y'-1 = (x-y)\left(xy'  + y\right)  \Rightarrow y'(1 -x^2 -xy) = 1 +xy -y^2

    So y''(1-x^2 -xy) + y'(-2x - xy' - y) =  xy' + y - 2yy' and hence using the fact that x=0, y=1, y' = 0 so y'' = 1?
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    (Original post by Zacken)
    \ln (y-x) = -xy \Rightarrow y'-1 = (x-y)\left(xy'  + y\right)  \Rightarrow y'(1 -x^2 -xy) = 1 +xy -y^2

    So y''(1-x^2 -xy) + y'(-2x - xy' - y) =  xy' + y - 2yy' and hence using the fact that x=0, y=1, y' = 0 so y'' = 1?
    ??
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    (Original post by Zacken)
    \ln (y-x) = -xy \Rightarrow y'-1 = (x-y)\left(xy'  + y\right)  \Rightarrow y'(1 -x^2 -xy) = 1 +xy -y^2

    So y''(1-x^2 -xy) + y'(-2x - xy' - y) =  xy' + y - 2yy' and hence using the fact that x=0, y=1, y' = 0 so y'' = 1?
    I got the same results. I didn't manipulate beforehand, just forced the differentiation through.
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    (Original post by EricPiphany)
    I got the same results. I didn't manipulate beforehand, just forced the differentiation through.
    Oh, phew! It's really ugly, what paper was that in?
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    (Original post by Zacken)
    Oh, phew! It's really ugly, what paper was that in?
    Not sure myself what this is, found the paper while browsing for extra FPn practice.
    http://papers.xtremepapers.com/CIE/C...1_s07_qp_1.pdf
 
 
 
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