Year 13 Maths Help Thread

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    (Original post by jamestg)
    Thanks! I'm sure I printed off this haha, probably stored it away with all my AS work...

    Looks like I'm going to have to get revising
    If you've done it right you won't need the formula book for any trig identities but it's nice to have. You'll be fine anyway from your AS results
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    Looks like I won't be wasting my frees this year, timetable comes on wednesday and I start thursday
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    How do you integrate  \surd (1-\sin x) given that  \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?
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    (Original post by Ano123)
    How do you integrate  \surd (1-\sin x) given that  \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?
    Use u=sin(x)
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    Why isn't the 4 negative?

    (Edexcel Jan 2006 C3 Q6a if you need to question)

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    (Original post by jamestg)
    Why isn't the 4 negative?

    (Edexcel Jan 2006 C3 Q6a if you need to question)
    It should be.. as far as I can see.
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    (Original post by jamestg)
    Why isn't the 4 negative?

    (Edexcel Jan 2006 C3 Q6a if you need to question)

    Ah I see, easy mistake to make you've expanded cos(x-a) rather than cos(x+a) given in the question.
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    (Original post by RDKGames)
    Use u=sin(x)
    What did you get?
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    (Original post by SeanFM)
    Ah I see, easy mistake to make you've expanded cos(x-a) rather than cos(x+a) given in the question.
    Ahh I see, thank you! I need to be more careful next time
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    (Original post by Ano123)
    What did you get?
    Is that restriction for x supposed to mean the limits of integration? I'm unsure how significant it is, but to integrate that expression;

    \displaystyle \int \sqrt{1-\text{sin}(x)} .dx

    let \displaystyle u=sin(x) \rightarrow du=cos(x) .dx

    \displaystyle \rightarrow \int \frac{\sqrt{1-u}}{cos(x)} .du

    Draw yourself a right angled triangle. Use the fact that u=sin(x) and you will find that cos(x)=\sqrt{1-u^2}

    \displaystyle \rightarrow \int \frac{\sqrt{1-u}}{\sqrt{1-u^2}} .du = \int \frac{\sqrt{1-u}}{\sqrt{(1+u)(1-u)}} .du

    and I'll let you finish that off.
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    (Original post by jamestg)
    Ahh I see, thank you! I need to be more careful next time
    No worries, you learn more from mistakes than getting things right, and eventually for each topic you build a mental checklist (trig is probably the worst for it) of things to look out for. Eg radians/degrees, cancelling solutions, right range of solutions etc. And if you have to, underlining key bits of information given in the question.
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    (Original post by Ano123)
    How do you integrate  \surd (1-\sin x) given that  \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?
    Note that  \sqrt{1-\sin x}\equiv \sqrt{sin^2(x/2) -2\sin (x/2)\cos (x/2) + \cos^2 (x/2)}
      \equiv \sqrt{(\sin (x/2)-\cos (x/2) )^2 } \equiv |\sin (x/2) - \cos (x/2)| .
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    (Original post by B_9710)
    This integral is not expected is C4 but you can still apply normal substitution methods.
    What cud i do
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    (Original post by youreanutter)
    What cud i do
    Exactly that. Use substitution x=\text{tan}\theta . The method itself is C3.

    If you're asking "How do I know to substitute that?" then just start with y=\text{arctan}(x) and find dy/dx. The substitution is based off this result.
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    (Original post by RDKGames)
    Exactly that. Use substitution x=\text{tan}\theta . The method itself is C3.

    If you're asking "How do I know to substitute that?" then just start with y=\text{arctan}(x) and find dy/dx. The substitution is based off this result.
    The substitution isn't based off that result. You could integrate 1/(1+x^2) independently without knowing the derivative of arctan x.
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    (Original post by B_9710)
    The substitution isn't based off that result. You could integrate 1/(1+x^2) independently without knowing the derivative of arctan x.
    As far as C3/C4 goes? How?
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    (Original post by RDKGames)
    As far as C3/C4 goes? How?
    It's not in C3/C4. But that doesn't change the methods for finding the integral.
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    I've cooked STEP I 2007 Q3; STEP I 2007 Q1 has almost finished cooking in the oven too.

    EDIT: STEP I 2007 Q1 ready too though I did some dodgy work when I ended up proving the general case in order to do show the special case.
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    Just joining this thread in advance now...

    I've dropped from Further Maths to just Maths, and I my applied unit will be M1...
    I'm sad to no longer be doing FMaths, but it didn't match my career aspirations, and there was no way I was taking 4 subjects :lol:

    I hate being in just a standard maths class already, though :lol:
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    STEP I 2007 Q8 in the bag too. Will try more MAT and IYGB papers now.
 
 
 
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