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    (Original post by nm786)
    Alright, then subbing in tan^2x=sec^2x-1 the integral now becomes \displaystyle \int \dfrac{4}{1- (sec^2x-1) } = \displaystyle \int \dfrac{4}{2-sec^2x} = \displaystyle \int-2cos^2x =\displaystyle \int -2(\frac{1}{2}cos2x+\frac{1}{2}) which then equals to \displaystyle \int -cos2x - 1 and i get \frac{1}{2}sin2x -x +C
    Okay, if you go back to the original expression that you had. On the bottom you had 1-{Sin^4(x)/Cos^4(x)} so what can you make this equal?
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    If you put it all over the same denominator you'd get (Cos^4(x)-Sin^4(x))/Cos^4(x) and can't this be expressed as Cos(2x)/Cos^4(x) So can you know simplify the integral by flipping this function and multiplying by the original numerator?
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    (Original post by MathsNerd1)
    Okay, if you go back to the original expression that you had. On the bottom you had 1-{Sin^4(x)/Cos^4(x)} so what can you make this equal?
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    If you put it all over the same denominator you'd get (Cos^4(x)-Sin^4(x))/Cos^4(x) and can't this be expressed as Cos(2x)/Cos^4(x) So can you know simplify the integral by flipping this function and multiplying by the original numerator?
    do you mean Cos^4(x)-Sin^4(x)/cos^4(x) be expressed as cos(4x)/cos^4(x)? I'm not sure why i have to flip this and multiply by the numerator sorry.
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    (Original post by nm786)
    do you mean Cos^4(x)-Sin^4(x)/cos^4(x) be expressed as cos(4x)/cos^4(x)? I'm not sure why i have to flip this and multiply by the numerator sorry.
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    Not exactly as if you remember your double angle formula you can't manipulate the formula in the way you've tried. Instead isn't the numerator the difference of 2 squares and as 1 of them is equal to one you'd get Cos(2x)/Cos^4(x) as stated

    Now remember that you still have 4Tan^2(x) over what we have just worked out, so how can that be simplified into a single fraction?

    If you want to see my method then look at this spoiler.
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    Name:  ImageUploadedByStudent Room1369946743.462397.jpg
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    I hope this has helped



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    (Original post by MedMed12)
    good good! how are they are going? I know you'll be OK
    Its been fine, just need to get more confidence with C3 again and build it up with the past papers for Maths. Been doing Chem and Bio questions alot this week.
    They are going fine dw lol
    I need to crack on with eco :/
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    how do you know whether to apply u-substitution or by parts when the question doesn't tell u anything?
    e.g. sind exact value of: 1/2 + 7/2 cos2θ
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    Can anyone give me C4 and C3 Jan 2013 please.
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    (Original post by tsr1)
    how do you know whether to apply u-substitution or by parts when the question doesn't tell u anything?
    e.g. sind exact value of: 1/2 + 7/2 cos2θ
    I'm guessing this is an integration question? If so, what are the limits? And you should be able to integrate:

    \dfrac{1}{2} + \dfrac{7}{2} \cos \2 \theta w.r.t. x

    without the need for substitution (it can be done very quickly in your head ).
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    if someone could integrate 1/sin^2x cos^2x id be grateful!!

    edit: scratch that, I've done it, was simpler than I thought...but feel free to give it a crack guys and girls
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    anyone know all the volume and surface area formulas we should know?
    thanks : )

    (I'm just doing connected rates of change C4 thats all)
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    (Original post by Namod)
    Can anyone give me C4 and C3 Jan 2013 please.
    they're both here with mark schemes

    http://mathspapers.co.uk/edexcel.html
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    (Original post by usycool1)
    I'm guessing this is an integration question? If so, what are the limits? And you should be able to integrate:

    \dfrac{1}{2} + \dfrac{7}{2} \cos \2 \theta w.r.t. x

    without the need for substitution (it can be done very quickly in your head ).
    Yes it is an integration question, it's easy that way but the question says use calculus to find the solution, also they have give 5 marks for using by-parts in the mark scheme.. It's june 2010 paper c4 , question 6 but I wouldn't have know what integration method to use...
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    Can't wait to start revising maths... ****ing Economics is painful.
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    (Original post by tsr1)
    Yes it is an integration question, it's easy that way but the question says use calculus to find the solution, also they have give 5 marks for using by-parts in the mark scheme.. It's june 2010 paper c4 , question 6 but I wouldn't have know what integration method to use...
    They were asking for,

    \displaystyle \int^{\pi /2}_0 \theta \left( \frac{1}{2} + \frac{7}{2} \cos 2\theta \right)  d\theta

    NOT,

    \displaystyle \int^{\pi /2}_0 \left( \frac{1}{2} + \frac{7}{2} \cos 2\theta \right)  d\theta
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    Hey guys do you have any tips on doing hard differentiation questions?

    Also, the C3 June 11 paper, q)5c) I am trying to differentiate M= 7.5e^-1/4ln3t to get dm/dt... and then make it = to -0.6ln3.

    I used the chain rule and did m=e^u and u=1/4ln3t and got dm/du=e^u and dm/dt= -1/12t.

    But I dont think im going along the right lines as I did this before and got the wrong answer.

    Any help is appreciated!
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    (Original post by pepeeglesfield)
    if someone could integrate 1/sin^2x cos^2x id be grateful!!

    edit: scratch that, I've done it, was simpler than I thought...but feel free to give it a crack guys and girls
    I got to the part where we had to integrate the integral of cosec2x dx.

    Do we have to know that the integral of cosec x = ln |cosecx - cotx| + c

    EDIT: Just found out that the integral is provided in the formula book
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    (Original post by frogs r everywhere)
    I got to the part where we had to integrate the integral of cosec2x dx.

    Do we have to know that the integral of cosec x = ln |cosecx - cotx| + c
    Yeah, but it's on the formula booklet.
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    (Original post by Story)
    Hey guys do you have any tips on doing hard differentiation questions?

    Also, the C3 June 11 paper, q)5c) I am trying to differentiate M= 7.5e^-1/4ln3t to get dm/dt... and then make it = to -0.6ln3.

    I used the chain rule and did m=e^u and u=1/4ln3t and got dm/du=e^u and dm/dt= -1/12t.

    But I dont think im going along the right lines as I did this before and got the wrong answer.

    Any help is appreciated!
    I presume you mean du/dt.

    Remember:

    u=-\dfrac{1}{4} \ln(3)t ; not u = -\dfrac{1}{4} \ln(3t).
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    (Original post by usycool1)
    I presume you mean du/dt.

    Remember:

    u=-\dfrac{1}{4} \ln(3)t ; not u = -\dfrac{1}{4} \ln(3t).
    So du/dt would be - 1/12 t with the t in the middle of the fraction?
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    Hey guys! Very sunny day out there! Gonna take a break in a few hours

    BUT Does anyone know how to do Solomon Press C3 Paper K Question 4 b
    Markscheme here: http://www.tomred.org/uploads/7/7/8/..._solomon_k.pdf

    Can someone please explain the justifications?


    (b) Use proof by contradiction to prove that log2 3 is irrational. (6)


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    (Original post by Better)
    Hey guys! Very sunny day out there! Gonna take a break in a few hours

    BUT Does anyone know how to do Solomon Press C3 Paper K Question 4 b
    Markscheme here: http://www.tomred.org/uploads/7/7/8/..._solomon_k.pdf

    Can someone please explain the justifications?


    (b) Use proof by contradiction to prove that log2 3 is irrational. (6)


    Remember you can express a rational number by a fraction, so you'll have a/b if it was rational and you've just got to go about proving that it isn't rational, therefore it's irrational.

    I wouldn't worry too much about this as I doubt you'd ever get asked this in the exam


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