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    In question 6! i cant remember the exact question but think it was:

    u = 4sin^2(x)

    prove = integral (4/(1-4x)^1/2 is same as (k integral sin^2(u)) where you must find k.

    the prove that the integral = k (integral) sin^2x i got 1/8 (integral) sec^2x.

    Where did i go wrong and how many marks would i have got for the follow up question if i used k= 1?8 instead of k =8?
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    (Original post by twigface)
    In the prove that the integral = k (integral) sin^2x i got 1/8 (integral) sec^2x.

    Where did i go wrong and how many marks would i have got for the follow up question if i used k= 1?8 instead of k =8?
    Your question doesn't make sense as it stands, please edit and clarify.
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    (Original post by twigface)
    In the prove that the integral = k (integral) sin^2x i got 1/8 (integral) sec^2x.

    Where did i go wrong and how many marks would i have got for the follow up question if i used k= 1?8 instead of k =8?
    you mean lambda not k right? all the method marks minus accuracy if you did it right
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    (Original post by WillRose)
    you mean lambda not k right? all the method marks minus accuracy if you did it right
    yeah i think so.

    do you remember what the question was?
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    (Original post by Zacken)
    Your question doesn't make sense as it stands, please edit and clarify.
    ok written out the question
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    (Original post by twigface)
    yeah i think so.

    do you remember what the question was?
    so probably 4/5?
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    (Original post by WillRose)
    you mean lambda not k right? all the method marks minus accuracy if you did it right
    actually refresh if you've seen this as i have edited
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    Not sure how you got sec^2T

    The 8 came from the dx/dT where T = theta
    I can't really remember the question but I remember something along the lines of tanx*8sinx*cosx I can't remember what was squared and what wasn't
 
 
 
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