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    (Original post by Zacken)
    Post your working then.
    I tried using chain rule this time;

    dy/dx = 4 + 4tan^2(4x)
    the 4 differentiates to 0 so i did chain rule of 4tan^2(4x);
    =4(tan^2(4x))
    =4[ (tan4x)^2 ]

    u= tan4x y = u^2
    du/dx = 4 + 4tan^2(4x) (as worked out in part a) dy/du = 2u (2tan4x)

    dy/dx = 4[ du/dx x dy/du]
    = 4[ (4 + 4tan^2(4x) ) x (2tan4x) ]
    = 4[ 8tan(4x) + 8tan^3(4x) ]
    = 32tan(4x) + 32tan^3(4x)
    = 32tan(4x) [ 1 + tan^2(4x) ]
    y = tan4x so;
    = 32y(1 + y^2)
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    (Original post by Abbas.13)
    I tried using chain rule this time;

    dy/dx = 4 + 4tan^2(4x)
    the 4 differentiates to 0 so i did chain rule of 4tan^2(4x);
    =4(tan^2(4x))
    =4[ (tan4x)^2 ]

    u= tan4x y = u^2
    du/dx = 4 + 4tan^2(4x) (as worked out in part a) dy/du = 2u (2tan4x)

    dy/dx = 4[ du/dx x dy/du]
    = 4[ (4 + 4tan^2(4x) ) x (2tan4x) ]
    = 4[ 8tan(4x) + 8tan^3(4x) ]
    = 32tan(4x) + 32tan^3(4x)
    = 32tan(4x) [ 1 + tan^2(4x) ]
    y = tan4x so;
    = 32y(1 + y^2)
    Yeah, that's correct, what's the problem?
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    (Original post by Zacken)
    Yeah, that's correct, what's the problem?
    I didn't get what you meant by use the rule tan^2(x) + 1 = sec^2(x)
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    (Original post by Abbas.13)
    I didn't get what you meant by use the rule tan^2(x) + 1 = sec^2(x)
    ...uhm, you used it several times in your answer.
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    (Original post by Zacken)
    ...uhm, you used it several times in your answer.
    I did? I don't recall making use of it??
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    June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead
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    (Original post by Fudge2)
    June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead
    Yeah I bloody hated it
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    (Original post by Fudge2)
    June 2013 made me lose the will to live. There's like 20 marks based on sketching/interpreting graphs. Give me 20 marks of integration instead
    Evaluate  \displaystyle \int_{1/4}^{1/2} \frac{1}{x^2\sqrt{ 1-x^2}} dx .
    May not be 20 marks. But it may be quite a few.
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    (Original post by Ano123)
    Evaluate  \displaystyle \int_{1/4}^{1/2} \frac{1}{x^2\sqrt{ 1-x^2}} dx .
    May not be 20 marks. But it may be quite a few.
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    How'd I do? cool question
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    (Original post by Fudge2)
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    How'd I do? cool question
    I got the same as you (2.14093) 2.14 to 3 s.f
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    (Original post by Fudge2)
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    How'd I do? cool question
    (Original post by xs4)
    I got the same as you (2.14093) 2.14 to 3 s.f
    Very good, but exact values are preferable - but I suppose it doesn't matter.
    I used u=1/x sub. Transforms it into a very easy integral. The exact value is 15 -3.
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    Hi,

    I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

    I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

    Solve:

    \int\frac{4+x}\sqrt{(4-2x^{2})}

    Giving your answer in the simplest rationalised form.

    I have worked out the solution and will post it tomorrow!
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    (Original post by Ano123)
    Very good, but exact values are preferable - but I suppose it doesn't matter.
    I used u=1/x sub. Transforms it into a very easy integral. The exact value is 15 -3.
    Yeah I tried to get an exact value but the lower limit of arcsin(0.25) isn't nice so my calculator wouldn't give me nice values u=1/x...I'd never spot that haha
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    (Original post by Fudge2)
    Yeah I tried to get an exact value but the lower limit of arcsin(0.25) isn't nice so my calculator wouldn't give me nice values u=1/x...I'd never spot that haha
    You could have still got an exact answer using your sub instead of mine. You just need to know what cot(arcsinx) is.
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    (Original post by Suits101)
    Hi,

    I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

    I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

    Solve:

    \int\frac{4+x}\sqrt{(4-2x^{2})}

    Giving your answer in the simplest rationalised form.

    I have worked out the solution and will post it tomorrow!
    Name:  Integral answer.jpg
Views: 240
Size:  499.9 KB
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    ∫ √(1+sin2x) dx
    This one is tricky.
    [latex doesn't seem to be working properly.]
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    (Original post by Suits101)
    Hi,

    I've literally just read the last page of this thread so I don't know if this question is considered difficult/easy so I might be posting inappropriately! And I've just had to look how to use LaTex...

    I don't know if these types of questions are still on spec, but I tried to make it difficult by combining quite a few rules and identities.

    Solve:

    \int\frac{4+x}\sqrt{(4-2x^{2})}

    Giving your answer in the simplest rationalised form.

    I have worked out the solution and will post it tomorrow!
    This would not come up on C3, only fp2
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    (Original post by Qcomber)
    This would not come up on C3, only fp2
    Hi.

    In the AQA textbook these questions are in the Core 3 section, especially since it's just a u substitution mixed with arcsin formula which the chapter includes.

    I haven't seen any come up, but just incase I posted it.
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    (Original post by Suits101)
    Hi.

    In the AQA textbook these questions are in the Core 3 section, especially since it's just a u substitution mixed with arcsin formula which the chapter includes.

    I haven't seen any come up, but just incase I posted it.
    If it does come up, they shouldn't have because knowledge of inverse trigonometrical integration is not on the spec for c3. Whereas the fp2 spec clearly states that you need to know that it even exists in the formula book.
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    (Original post by Ano123)
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    yep thats what i got. I'm guessing you're doing fp2
 
 
 
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