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    (3i) , so as usual, where did I mess up?
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    (Original post by ckfeister)
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    (3i) , so as usual, where did I mess up?
    You had \displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1, not \displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1 (as you treated it)

    Instead, divide both sides by 1+y^2, to obtain,

    \displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1. Should be relatively clear where this is going.
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    (Original post by ckfeister)
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    (3i) , so as usual, where did I mess up?
    To be honest, I'm struggling to make any sense of that at all. What does the (x dx) (-y^2) bit mean?

    The first step is to separate the variables i.e. formally write:

    \frac{dy}{1+y^2} = dx

    then integrate.
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    (Original post by _gcx)
    You had \displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1, not \displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1 (as you treated it)

    Instead, divide both sides by 1+y^2, to obtain,

    \displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1. Should be relatively clear where this is going.
    I don't get where this is meant to go, whats the link with tanx?
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    (Original post by ckfeister)
    I don't get where this is meant to go, whats the link with tanx?
    Are you familiar with how to approach

    \displaystyle \int \frac 1 {1+y^2} \mathrm dy?
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    (Original post by _gcx)
    Are you familiar with how to approach

    \displaystyle \int \frac 1 {1+y^2} \mathrm dy?
    Nope... first time. I would try sub but I'm sure it won't get tan(x)
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    (Original post by ckfeister)
    Nope... first time.
    Have you covered integration by substitution? Try substituting y = \tan\theta.
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    (Original post by ckfeister)
    I don't get where this is meant to go, whats the link with tanx?
    Which module are you doing currently?
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    (Original post by notnek)
    which module are you doing currently?
    aqa - fp2
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    (Original post by ckfeister)
    aqa - fp2
    Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

    Do you have any ideas which substitution you can use?
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    (Original post by Notnek)
    Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

    Do you have any ideas which substitution you can use?
    x/a tan-1 x/a?
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    (Original post by ckfeister)
    u = 1+y^2

    du/dy= 2y

    du/2y = dy

    Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.
    It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.
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    (Original post by ckfeister)
    u = 1+y^2

    du/dy= 2y

    du/2y = dy

    Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.
    Have you covered integration by substitution? Try substituting y = \tan\theta.
    It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.
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    (Original post by Notnek)
    It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.
    (Original post by _gcx)
    It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.
    Is it x/a tan-1 x/a?
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    (Original post by ckfeister)
    Is it x/a tan-1 x/a?
    Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)
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    (Original post by _gcx)
    Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)
    C3 - which I learnt there, I'm going to watch khan academy (£350 course cost doesn't cover this)
 
 
 
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