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    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
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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
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    \displaystyle \int \sec x dx=\int \dfrac{1}{\cos x} dx =\int \dfrac{cosx}{cos^2x}dx = \int \dfrac{cosx}{1-sin^2x} dx

    \displaystyle u=sinx \Rightarrow du=dxcosx

    \displaystyle I=\int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \int \dfrac{1}{u + 1} - \dfrac{1}{u - 1}du

    \displaystyle =\dfrac{1}{2} \left( ln(u+1)-ln(u-1) \right )+C=\dfrac{1}{2}ln\dfrac{sinx+1}  {sinx-1}=\frac {1}{2}\ln\frac {(1+\sin x)^2}{\cos^2 x}+C
    \displaystyle =ln(secx+tanx)+C
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    (Original post by Kvothe the arcane)
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    \displaystyle \int \sec x dx=\int \dfrac{1}{\cos x} dx =\int \dfrac{cosx}{cos^2x}dx = \int \dfrac{cosx}{1-sin^2x} dx
    \displaystyle u=sinx \Rightarrow du=dxcosx

    \displaystyle I=\int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \int \dfrac{1}{u + 1} - \dfrac{1}{u - 1}du

    \displaystyle =\dfrac{1}{2} \left( ln(u+1)-ln(u-1) \right )+C=\dfrac{1}{2}\dfrac{ln(sinx+1  )}{ln(sinx-1)}=\frac {1}{2}\ln\frac {(1+\sin x)^2}{\cos^2 x}+C
    \displaystyle =ln(secx+tanx)+C
    Perfect! Spoiler tag please
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    (Original post by aymanzayedmannan)
    Perfect! Spoiler tag please
    Done .

    The intended approach is so unintuitive. I have a spare few minutes at work so might see if I can do it another way.
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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
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    Obviously wouldn't have been able to think of this without knowing the result already; without further ado: u = \sec x + \tan x

    \displaystyle

\begin{equation*} \int \sec x \, \mathrm{d}x = \int \sec x\frac{\sec x + \tan x}{\sec x + \tan x} \, \mathrm{d}x = \int \frac{\mathrm{d}u}{u}=\ln (\sec x + \tan x) + \mathcal{C}\end{equation*}
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    (Original post by Kvothe the arcane)
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    \displaystyle \int \sec x dx=\int \dfrac{1}{\cos x} dx =\int \dfrac{cosx}{cos^2x}dx = \int \dfrac{cosx}{1-sin^2x} dx

    \displaystyle u=sinx \Rightarrow du=dxcosx

    \displaystyle I=\int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \int \dfrac{1}{u + 1} - \dfrac{1}{u - 1}du

    \displaystyle =\dfrac{1}{2} \left( ln(u+1)-ln(u-1) \right )+C=\dfrac{1}{2}\dfrac{ln(sinx+1  )}{ln(sinx-1)}=\frac {1}{2}\ln\frac {(1+\sin x)^2}{\cos^2 x}+C
    \displaystyle =ln(secx+tanx)+C
    That looks ****ing terrifying
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    (Original post by Dohaeris)
    (Original post by Kvothe the arcane)
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    \displaystyle \int \sec x dx=\int \dfrac{1}{\cos x} dx =\int \dfrac{cosx}{cos^2x}dx = \int \dfrac{cosx}{1-sin^2x} dx

    \displaystyle u=sinx \Rightarrow du=dxcosx

    \displaystyle I=\int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \int \dfrac{1}{u + 1} - \dfrac{1}{u - 1}du

    \displaystyle =\dfrac{1}{2} \left( ln(u+1)-ln(u-1) \right )+C=\dfrac{1}{2}\dfrac{ln(sinx+1  )}{ln(sinx-1)}=\frac {1}{2}\ln\frac {(1+\sin x)^2}{\cos^2 x}+C
    \displaystyle =ln(secx+tanx)+C
    That looks ****ing terrifying
    Hey, how so?
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    (Original post by Kvothe the arcane)
    Hey, how so?
    ..
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    (Original post by EricPiphany)
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    Pardon? I couldn't think of another solution. I just ended up with mammoth things which were hard to solve.
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    (Original post by Kvothe the arcane)
    Pardon? I couldn't think of another solution. I just ended up with mammoth things which were hard to solve.
    I joke.
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    (Original post by Kvothe the arcane)
    Hey, how so?
    I don't understand it, so I'm afraid of it

    Seriously though, kinda ingenious. Not something I would ever be able to do, unfortunately
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    (Original post by Dohaeris)
    I don't understand it, so I'm afraid of it

    Seriously though, kinda ingenious. Not something I would ever be able to do, unfortunately
    Oh, what part? I recognised secx=1/cosx and cosx\cos^2x and used a trig identity.
    I then used a substitution.
    And then I integrated, resubbed and then did some manipulation.
    It's nothing otherworldly and I'm sure you'll be able to do the same thing once you understand .
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    (Original post by Kvothe the arcane)
    Oh, what part? I recognised secx=1/cosx and cosx\cos^2x and used a trig identity.
    I then used a substitution.
    And then I integrated, resubbed and then did some manipulation.
    It's nothing otherworldly and I'm sure you'll be able to do the same thing once you understand .
    The thing is, looking at the answer, I completely understand it. The problem, as it so often is, is that it comes too late lol. I could look at that question for an hour before deciding that it's some uni ****, but when I look at the answer I slap myself silly for not doing the simpler stuff at least.

    This does look like one of those question that are a bit above C4 level, though using C4 material, isn't it? I suppose that's why the answer for it is given in the formula booklet
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    (Original post by Kvothe the arcane)
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    \displaystyle \int \sec x dx=\int \dfrac{1}{\cos x} dx =\int \dfrac{cosx}{cos^2x}dx = \int \dfrac{cosx}{1-sin^2x} dx

    \displaystyle u=sinx \Rightarrow du=dxcosx

    \displaystyle I=\int \dfrac{1}{1-u^2} du = \dfrac{1}{2} \int \dfrac{1}{u + 1} - \dfrac{1}{u - 1}du

    \displaystyle =\dfrac{1}{2} \left( ln(u+1)-ln(u-1) \right )+C=\dfrac{1}{2}\dfrac{ln(sinx+1  )}{ln(sinx-1)}=\frac {1}{2}\ln\frac {(1+\sin x)^2}{\cos^2 x}+C
    \displaystyle =ln(secx+tanx)+C
    One qualm,

    \displaystyle \frac{\log(\sin(x)+1)}{\log(\sin  (x)-1)} \neq \log\left(\frac{\sin(x)+1}{\sin(  x)-1}\right)

    Though that was a typo, nicely done
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    (Original post by Euclidean)
    One qualm,

    \displaystyle \frac{\log(\sin(x)+1)}{\log(\sin  (x)-1)} \neq \log(\frac{\sin(x)+1}{\sin(x)-1})

    Though that was a typo, nicely done
    I've amended but thanks .
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    (Original post by Euclidean)
    ...
    \LaTeX pro tips: to render your brackets correctly with fractions and the like, use \left( and \right)

    (Original post by Kvothe the arcane)
    I've amended but thanks .
    As above; use \cos, \ln, \sin so that you're typesetting functions instead of squashing the variables c, o and s together.
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    (Original post by Zacken)
    \LaTeX pro tips: to render your brackets correctly with fractions and the like, use \left( and \right)



    As above; use \cos, \ln, \sin so that you're typesetting functions instead of squashing the variables c, o and s together.
    Thanks. I always forget to use \ before such functions. Though my lack of proper editing was due to time constraints. I may go back and add aligns etc.
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    (Original post by Zacken)
    \LaTeX pro tips: to render your brackets correctly with fractions and the like, use \left( and \right)


    Thanks though will do
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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
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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
    Wow that's a really good idea
 
 
 
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