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    After slogging my through the first two parts of the question in a whole hour , I have become stuck at part c. Its worth 9 marks and I just find a way to Integrate this.

    I have tried the sub u = cos x, and got to Integrate (√(2u^2 - 1) du). ANd it certianly doesn't work by parts.

    Could some one help please?

    Thanks,
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    Is this part of the July 2005 AEA paper? If so, you can use the substitution u=sinx and then use earlier results to evaluate the integral.
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    okay, this is what i did, it may be wrong though... (i'm sure i've done somthing wrong to be honest but this is what i would have done)

    sin2x ≡ 2sinxcosx so it follows that

    sin2x/2cosx ≡ sinx

    so rewrite the integral

    ∫ (sin2x/2cos2x) . √cos2x dx

    then

    ∫(sin2x/2√cos2x)dx

    then let u = √cos2x

    so that du/dx = -sin2x/√cos2x which is actually -sin2x/u

    then you have that du = (-sin2x/u)dx

    and

    -1/2du = (sin2x/2u) dx

    going back to the original integral:

    substitute u for √cos2x

    ∫(sinx/2u) dx

    and thus

    ∫ -1/2 du

    or -1/2∫1du

    now to convert limits which are in x, u = √cos 2(π/4) = 0 and u = √cos 2(0) = 1

    then you have an integral:

    limits 0, 1

    -1/2∫1du = -1/2{[u]}

    = -1/2 {[1] - [0]}

    = -1/2
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    I think you make a mistake from
    sin(2x)/2cosx = sinx and inserting it into integral because u make
    cosx - ->cos(2x).
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    im not really sure about this but u no u can rewrite cos 2x as cos^2(x)-1. so can u see what to do next?
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    so can u see what to do next?

    No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

    Gaz, how do you mean, use part of the question (earlier results)?
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    aha!, you're right, i knew that solution was too simple. thanks for pointing that out.
    i am smacking my head right now... that is quite a big mistake (i can't believe i did that!!!)
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    (Original post by Vazzyb)
    so can u see what to do next?

    No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

    Gaz, how do you mean, use part of the question (earlier results)?
    lol sorry im wrong just ignore me...
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    (Original post by Vazzyb)
    so can u see what to do next?

    No lol. Thats exactly what I did, to get root of (2u^2 - 1) du..

    Gaz, how do you mean, use part of the question (earlier results)?
    I remembered doing this question in the exam and I'm fairly sure I used the earlier parts. I can't remember the specifics though so people might be able to help better if you posted the entire question.
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    OK cheers, The whole question...

    [Attached]

    (sorry about not doing it earlier!)
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    (Original post by Gaz031)
    I remembered doing this question in the exam and I'm fairly sure I used the earlier parts. I can't remember the specifics though so people might be able to help better if you posted the entire question.
    Well after gaz's clue i think you have to rewrite cos2x as 1- 2sin^2x
    so ∫sinx√cos2x=∫sinx√(1- 2sin^2x)
    Then i would be able to do it by recognition because im not very good with substitution... Do u guys know any resources that help with integration by substitution?

    Edit:sorry im wrong again lol please do ignore me
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    Looking at it, I think there's a fairly straight forward plan of attack:

    I = \int \sqrt{2u^2-1}\, du = u\sqrt{2u^2-1} - \int \frac{2u^2}{\sqrt{2u^2-1}} \\

= u\sqrt{2u^2-1} - I - \int \frac{1}{\sqrt{2u^2-1}} \,du \\

= u\sqrt{2u^2-1} - I - \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{y^2-1}} \,dy \\

= u\sqrt{2u^2-1} - I - \frac{1}{\sqrt{2}} \cosh ^{-1}(y)

    Where I've used y2=2u2. Rearranging gives:

    I = \frac{1}{4}[2u\sqrt{2u^2-1} - \sqrt{2} \cosh ^{-1}(\sqrt{2}u)]
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    (Original post by Wrangler)
    Looking at it, I think there's a fairly straight forward plan of attack:

    I = \int \sqrt{2u^2-1}\, du = u\sqrt{2u^2-1} - \int \frac{2u^2}{\sqrt{2u^2-1}} \\

= u\sqrt{2u^2-1} - I - \int \frac{1}{\sqrt{2u^2-1}} \,du \\

= u\sqrt{2u^2-1} - I - \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{y^2-1}} \,dy \\

= u\sqrt{2u^2-1} - I - \frac{1}{\sqrt{2}} \cosh ^{-1}(y)

    Where I've used y2=2u2. Rearranging gives:

    I = \frac{1}{4}[2u\sqrt{2u^2-1} - \frac{1}{\sqrt{2}} \cosh ^{-1}(\sqrt{2}u)]
    can't be right, since this paper only requires c1-c4 knowledge.
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    (Original post by Widowmaker)
    can't be right, since this paper only requires c1-c4 knowledge.
    Well, I'm pretty sure that's right. It may not be the way you would have gone about it.
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    have u tried using the substitution u = √2cosx?
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    i think i got it:

    cos2x = 2cos²x - 1

    so ∫sinx√(2cos²x -1) dx

    then let u = √2cosx

    then du/dx = -√2sinx

    then you get:

    -√2∫√u²-1 du with limits when x = π/4 u = 1, and when x = 0, u = √2

    then make substitution u = sec θ,

    giving (from (a)) ∫secθtan²θdθ with limits, θ = π/4 and θ = 0

    then if you did part (b), you just substitute in and you're done!

    --------------

    HEY!, that was my idea
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    (Original post by Mr. Pink)
    i think i got it:

    cos2x = 2cos²x - 1

    so ∫sinx√(2cos²x -1) dx

    then let u = √2cosx

    then du/dx = -√2sinx

    then you get:

    -√2∫√u²-1 du with limits when x = π/4 u = 1, and when x = 0, u = √2

    then make substitution u = sec θ,

    giving (from (a)) ∫secθtan²θdθ with limits, θ = π/4 and θ = 0

    then if you did part (b), you just substitute in and you're done!

    --------------

    HEY!, that was my idea
    lol i guess it was, i was about to do it but i suck at substitution
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    i'm glad i figured it out though, that other mistake i made before was REALLY bad
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    (Original post by Mr. Pink)
    i'm glad i figured it out though, that other mistake i made before was REALLY bad
    yeah i made quite a few mistakes as well, i thought the derivative of sin^2x was 2 sinx lol. I need more sleep....
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    edit : nevermind eh? :p:
 
 
 
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