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    (Original post by Jackpalmer1996)
    Don't worry if you haven't and others have, there's still plenty of time until the exam! Loads of cramming to do
    Best of luck!!!

    Thank you 😄 you just made me feel better
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    Good luck guys
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    Any new C3/4 crash maths papers yet ?
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    hey I need help with JAN IAL C34 question 4 im stuck on it
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    Is there a thread for C3???
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    and how is everyone revising for maths? just practice? is that the only thing i can do?
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    (Original post by imran_)
    hey I need help with JAN IAL C34 question 4 im stuck on it
    yeah?
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    (Original post by Zacken)
    yeah?
    sorry, idk how to type it on here but heres the link

    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    its question 4
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    (Original post by SSD07)
    Is there a thread for C3???
    http://www.thestudentroom.co.uk/show...5#post63691635
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    (Original post by SSD07)
    Is there a thread for C3???
    http://www.thestudentroom.co.uk/show....php?t=3847081
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    (Original post by imran_)
    sorry, idk how to type it on here but heres the link

    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    its question 4
    What are you stuck on? the limits becomes 0 \, and \frac{\pi}{3} your \frac{\mathrm{d}x}{\mathrm{d}u} = \cos \theta so your integral is transformed into:

    \displaystyle 

\begin{equation*}\int_0^{\pi/3} \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}} \, \mathrm{d}\theta\end{equation*}

    Now just use the fact that \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cdots
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    (Original post by Zacken)
    What are you stuck on? the limits becomes 0 \, and \frac{\pi}{3} your \frac{\mathrm{d}x}{\mathrm{d}u} = \cos \theta so your integral is transformed into:

    \displaystyle 

\begin{equation*}\int_0^{\pi/3} \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}} \, \mathrm{d}\theta\end{equation*}

    Now just use the fact that \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cdots
    what happens to the power 3/2
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    (Original post by imran_)
    what happens to the power 3/2
    Aye, sorry I mis-read that as \frac{1}{2} because I'm blind. We have that (1-\sin^2 \theta)^{3/2} = \sqrt{1-\sin^2 \theta}(1- \sin^2 \theta) by indices laws, so your integrand is:

    \displaystyle \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}(1-\sin^2 \theta)} = \frac{\cos \theta}{2\cos \theta \cos^2 \theta}
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    thanks! what about s2?
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    (Original post by SSD07)
    and how is everyone revising for maths? just practice? is that the only thing i can do?
    Pretty much all I'm doing, just loads and loads of past papers and exam questions, the solomon papers are great for the tougher questions if you're aiming for A/A* !

    http://www.physicsandmathstutor.com/...rs/c4-solomon/

    Good luck!
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    Is anyone else finding the Solomon papers for c3 easier than the actual edexcel pprs


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    Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?
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    (Original post by Jacob2215)
    Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?
    Link the paper? Cba to go find it rn
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    (Original post by Jacob2215)
    Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?
    Have you seperated the variables?
    Put the functions of t on one side(right side preferably) and N on Left side.


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    (Original post by physicsmaths)
    Have you seperated the variables?
    Put the functions of t on one side(right side preferably) and N on Left side.


    Posted from TSR Mobile
    Yeah I think i see what I'm doing now, cheers cheers
 
 
 
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