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    Can anyone help me with this question? Name:  IMG_20160329_010916.png
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    (Original post by RUNSran)
    Can anyone help me with this question? Name:  IMG_20160329_010916.png
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    What have you done so far?
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    (Original post by undercxver)
    What have you done so far for part c?
    i actually do not know how to start or where to start from
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    (Original post by RUNSran)
    Can anyone help me with this question? Name:  IMG_20160329_010916.png
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    You know that the area is \displaystyle \int_a^b y \, \mathrm{d}x = \int_a^b y \, \frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d}t

    So you can replace y and \frac{\mathrm{d}x}{\mathrm{d}t} by their respective expressions and the limits should follow from from x=0 at which you need to find the value of t that makes this so and then do the same for the other limit (which you've cropped out of the question, so...)
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    (Original post by RUNSran)
    i actually do not know how to start or where to start from
    You've been given parametric equations, you can make it \frac{dy}{dx} by differentiating x=cos2t and y=cosect.

    Anyways, I'm assuming you should've done part a and b.

    Find what t equals when x=0 .

    This will be your limits for when you integrate to find the shaded area.

    I hope this helps you start of?
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    (Original post by Zacken)
    You know that the area is \displaystyle \int_a^b y \, \mathrm{d}x = \int_a^b y \, \frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d}t

    So you can replace y and \frac{\mathrm{d}x}{\mathrm{d}t} by their respective expressions and the limits should follow from from x=0 at which you need to find the value of t that makes this so and then do the same for the other limit (which you've cropped out of the question, so...)
    Oh, I get it - Thanks mate
    what about 6 a? I tried solving it by parts but it seems I am missing something Name:  IMG_20160329_012520.png
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    (Original post by undercxver)
    You've been given parametric equations, you can make it \frac{dy}{dx} by differentiating x=cos2t and y=cosect.

    Anyways, I'm assuming you should've done part a and b.

    Find what t equals when x=0 .

    This will be your limits for when you integrate to find the shaded area.

    I hope this helps you start of?
    Yeah I figured out how to obtain the limits in terms of t but getting the integral terms was the challenging bit - someone has explain it to me.
    i have posted posted another question and I will be very glad if you could help
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    (Original post by RUNSran)
    Oh, I get it - Thanks mate
    what about 6 a? I tried solving it by parts but it seems I am missing something Name:  IMG_20160329_012520.png
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    I tried integration by parts too and got .

    Doesn't look right though. :s: is another answer I got. :lol:

    I'm sorry but I'm not great at integration by parts yet. Maybe the way you did it is wrong? What exactly did you do step by step when integrating?
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    (Original post by RUNSran)
    Oh, I get it - Thanks mate
    what about 6 a? I tried solving it by parts but it seems I am missing something Name:  IMG_20160329_012520.png
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    expand cos(A+B) - cos(A-B) and see if that helps


    dont open if youre doing step...
    Spoiler:
    Show
    if you want more practice, try Q7, STEP I, 2009
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    (Original post by RUNSran)
    Oh, I get it - Thanks mate
    what about 6 a? I tried solving it by parts but it seems I am missing something Name:  IMG_20160329_012520.png
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    'Factor formula' -  \displaystyle \cos A - \cos B = -2\sin \left (\frac{A+B}{2} \right )\sin \left (\frac{A-B}{2} \right ) .
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    Thank very much guys - I will try using the using the suggested Identities and see.
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    (Original post by undercxver)
    I tried integration by parts too and got .

    Doesn't look right though. :s: is another answer I got. :lol:

    I'm sorry but I'm not great at integration by parts yet. Maybe the way you did it is wrong? What exactly did you do step by step when integrating?
    I somehow can't see the pictures you uploaded .
    This is what I did..Name:  IMG_20160329_081335.jpg
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