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    (Original post by PythianLegume)
    With integration and differentiation, it's generally radians. Also, think about the values in that question - if it asks about 2 weeks, when t=2, you're not going to be doing sin of 1 degree.
    Thanks, didnt think of it like that.
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    (Original post by suncake)
    Say you have something like this... Can you just take out 1/3 as a factor and put it outside the integral?
    Attachment 227729

    Posted from TSR Mobile
    Yes, I would do that to simplify the question.


    (Original post by PhoenixSeeker)
    I still can't get it to work is it possible to do it this way?
    I'm fairly sure you can't do it that way with C4 techniques.
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    (Original post by Story)
    Yeah. I did 1/cos^2x = 1/(1/2cos2x+1), then when flipping it I think I made an error as I wrote 2/2cosx+1..but does the half become a 2 or is it just 1/2....Is it 2(cos2x+1)..?
    Firstly, cos^2(x) = 1/2cos2x + 1/2! But I don't get how you're flipping it?
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    (Original post by orange94)
    You mean like this ?

    Attachment 227728


    Posted from TSR Mobile
    As Maths 247 put it,

    Let them get married! ( i dont think you have done anything wrong?)
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    (Original post by suncake)
    Say you have something like this... Can you just take out 1/3 as a factor and put it outside the integral?
    Attachment 227729

    Posted from TSR Mobile
    \frac{2}{3}\int \frac{1}{2x-1} - \frac{1}{3}\int \frac{1}{x+1}
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    (Original post by justinawe)
    3^2^x = (3^2)^x = 9^x
    (Original post by PythianLegume)
    Well y=32x

    ln(y) = 2xln(3)
    (1/y)(dy/dx)=2ln(3)
    dy/dx=2 x 32xln(3)

    Hence integration divides by 2ln(3).
    Ohhh I get it - thanks! Is this in the syllabus btw? Haven't come across it much before :s
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    (Original post by Myocardium)
    Ohhh I get it - thanks! Is this in the syllabus btw? Haven't come across it much before :s
    The differential of a^x is definitely in the syllabus.
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    (Original post by frogs r everywhere)
    Reminds me of good ol' GCSE days.

    Pretty colours everywhere.
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    Thank you so much!
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    (Original post by Kardy)
    Form simultaneous equations to find lamda. The two equations can be formed from the dot product of l1 & AP (which are perpendicular so = 0 ) and the other equation you already have, it's the equation of the line l1.

    http://www.examsolutions.net/a-level...7&solution=7.3

    Thanks Kardy so much..!!!!!!!!!
    Good luck tomorrow...!!!!!
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    Welcome Squad
    (Original post by Myocardium)
    How would you integrate something like 3^x?
    This is the 3rd time this has been asked today ( I asked first), everyone will be able to do this in their sleep now, ha ha just remember when you differentiate you get 3^x.ln3 but when you integrate you get 3^x/ln3

    EDIT: Don't ever forget the +c
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    (Original post by frogs r everywhere)
    Multiply by k, Take exponentials, Get rid of the denominator by multiplying through, Expand brackets and take it from there.

    but I have two k's though so I can't multiply out until I have taken e to both sides...
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    (Original post by Westeros)
    \frac{2}{3}\int \frac{1}{2x-1} - \frac{1}{3}\int \frac{1}{x+1}
    I think you can. Why would you not just take 1/3 out?
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    (Original post by suncake)
    Say you have something like this... Can you just take out 1/3 as a factor and put it outside the integral?
    Attachment 227729


    Actually, could someone work through this for me Death by lns...

    Posted from TSR Mobile
    Well I did! You right!

    Unless anyone here objects to 1/3 being pulled out.

    Here is what I did

    Name:  ImageUploadedByStudent Room1371492088.213893.jpg
Views: 155
Size:  144.2 KB


    Posted from TSR Mobile
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    (Original post by Myocardium)
    Ohhh I get it - thanks! Is this in the syllabus btw? Haven't come across it much before :s
    Page 41 in your Edexcel C4 textbook

    (Original post by Supes180)
    I think you can. Why would you not just take 1/3 out?
    Hmm, I don't know. I'd just approach it this way because it's easier for me

    (Original post by lefterispower)
    Yeaa i know that...but i dont know the steps before doing a.b=0

    if you have time to write me the solution i would be so happy...
    https://www.youtube.com/watch?v=ROC5tCZWPRs This video explains it all!
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    (Original post by imasha.sj)
    Thank you but why use the two t values for x=2.5 instead of using the t values for x=2 and x=2.5?
    I didnt try to find the t values for x. did you? i dont think there is a suitable value :/ It made sense when i drew a vertical line at x=2.5 and marked the points where x=2.5 cuts the curve ( upper point being t= 2 and lower t=.5 ) As long as the area is bounded by values you are aware of, you can integrate! Over here we used t, because we had already found its values. I didnt really try changing x= 1 and x=2.5 to t. too much working :P
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    (Original post by orange94)
    Well I did! You right!

    Unless anyone here objects to 1/3 being pulled out.

    Here is what I did

    Name:  ImageUploadedByStudent Room1371492088.213893.jpg
Views: 155
Size:  144.2 KB


    Posted from TSR Mobile
    Integral of 2/(2x-1) you wrote as ln|2x-1| is this correct?
    Also don't forget +C on your second line
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    I can't believe after doing hundreds of integration questions I still manage to completely forget about the c argg
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    I've no idea how to draw graphs from parametric equations. Eek
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    (Original post by Westeros)
    Integral of 2/(2x-1) you wrote as ln|2x-1| is this correct?
    Do you think it isn't? One of us is missing something here.
 
 
 
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