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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by BBeyond)
    Fair probs should have guessed Camb?
    He's not that good.
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    These problems are getting somewhat above c4 standard..
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    (Original post by 13 1 20 8 42)
    These problems are getting somewhat above c4 standard..
    :teehee:
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    (Original post by Zacken)
    3. Using \theta \mapsto \frac{\pi}{4} - \theta, evaluate: \displaystyle \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta.

    Extension: Evaluate \displaystyle \int_0^{1} \frac{\ln (1 + x)}{1+x^2} \, \mathrm{d}x.
    (Original post by Zacken)
    4. Evaluate \displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

    5. Generalise this substitution and prove that it holds for all smoothly behaved f when evaluating \displaystyle \int_a^b f(x) \, \mathrm{d}x
    Saving these beauties for later
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    (Original post by Euclidean)
    Saving these beauties for later
    Let me know how you get on. :yep:
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    (Original post by Zacken)
    4. Evaluate \displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

    5. Generalise this substitution and prove that it holds for all smoothly behaved f when evaluating \displaystyle \int_a^b f(x) \, \mathrm{d}x
    Ffs misread your first question here and have been trying to integrate ln(1+sinx) instead and getting absolutely nowhere
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    (Original post by Zacken)
    If I say x \mapsto f(x), I'm really just making a substitution u = f(x) then u=x. :-)

    Well if you call your integral I - then you have I = \ln 2 - I.

    (although that's not quite correct, use that idea).
    I've never seen that before. I still don't know what to do though. If I let the original integral = I and mine = J, I have that J = the integral of ln(2) - I? Is that the right idea?
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    (Original post by Zacken)
    :teehee:
    I just hope if any non-STEPers/maths students/general maths enthusiasts see them they don't get too scared..

    I'll throw in a standard that everyone should know to level the balance

    \displaystyle \int \sqrt{1 + cosx} dx
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    (Original post by Fudge2)
    I've never seen that before. I still don't know what to do though. If I let the original integral = I and mine = J, I have that J = the integral of ln(2) - I? Is that the right idea?
    Don't worry about not seeing it before, it's a little advanced.

    If we let the original integral be I = \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta, then our substitution gives us:

    \displaystyle I = \int_0^{\pi/4} \ln 2 - \ln (1 + \tan\theta) \, \mathrm{d}\theta = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - \int_0^{\pi/4} \ln(1+\tan\theta) \, \mathrm{d}\theta

    Which is just: I = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - I, so you can re-arrange and solve for I.
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    (Original post by BBeyond)
    Ffs misread your first question here and have been trying to integrate ln(1+sinx) instead and getting absolutely nowhere
    Yeah, the mis-read one is a hard one. The definite integral is given in terms of Catalan's constant: 2\mathcal{C} - \frac{\pi}{2}\ln 2. :teehee:
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    (Original post by Zacken)
    Don't worry about not seeing it before, it's a little advanced.

    If we let the original integral be I = \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta, then our substitution gives us:

    \displaystyle I = \int_0^{\pi/4} \ln 2 - \ln (1 + \tan\theta) \, \mathrm{d}\theta = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - \int_0^{\pi/4} \ln(1+\tan\theta) \, \mathrm{d}\theta

    Which is just: I = \int_0^{\pi/4} \ln 2 \, \mathrm{d}\theta - I, so you can re-arrange and solve for I.
    Haha I was kinda hoping that was the case! That's really neat though. I got pi/8ln2...?
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    (Original post by Fudge2)
    Haha I was kinda hoping that was the case! That's really neat though. I got pi/8ln2...?
    Yeah, nothing like this would ever come up on a C4 paper. That's correct, good work!
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    (Original post by Zacken)
    Yeah, nothing like this would ever come up on a C4 paper. That's correct, good work!
    Wooo! Thanks for your guidance
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    (Original post by Fudge2)
    Wooo! Thanks for your guidance
    Enjoyed it?
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    (Original post by Zacken)
    4. Evaluate \displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

    5. Generalise this substitution and prove that it holds for all smoothly behaved f when evaluating \displaystyle \int_a^b f(x) \, \mathrm{d}x
    Spoiler:
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    Name:  ImageUploadedByStudent Room1456765243.094154.jpg
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    Wasn't so sure about this one but I got -pi/2(ln2) ? I attached my solution as I'm almost sure this is wrong
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    (Original post by Zacken)
    Enjoyed it?
    Yeah! Despite the fact I'll probably never have to use that trick again it was cool...
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    (Original post by 13 1 20 8 42)
    \displaystyle \int \sqrt{1 + cosx} dx
    2\sqrt{2}sin\frac{x}{2} + C

    Spoiler:
    Show
    cos2x = 2cos^{2}x-1

    cosx = 2cos^{2}(\frac{x}{2})-1

    \therefore 1+cosx = 2cos^{2}(\frac{x}{2})

    \displaystyle \int \sqrt{2cos^{2}(\dfrac{x}{2})}\ dx \ \rightarrow\ \sqrt{2}\displaystyle \int \sqrt{cos^{2}(\dfrac{x}{2})}\ dx

    \sqrt{2} \times \dfrac{1}{0.5}sin(\dfrac{x}{2}) + C

    (\dfrac{1}{\frac{1}{2}} doesn't look nice lmao)
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    (Original post by BBeyond)
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    Wasn't so sure about this one but I got -pi/2(ln2) ? I attached my solution as I'm almost sure this is wrong
    Posted from TSR Mobile
    Yep, that's fine. But unwieldy but it works. Now try doing the general case.
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    (Original post by Fudge2)
    Yeah! Despite the fact I'll probably never have to use that trick again it was cool...
    I've used it a hundred times over when doing integration problems, not at C4 level, but you'll see. :-)
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    (Original post by Zacken)
    Yep, that's fine. But unwieldy but it works. Now try doing the general case.
    Is there a quicker way of doing that? Buzzing with that though ahah didn't even know about this trick until today. I think that might be a bit past my level lol
 
 
 
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