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    I'm really stuck on these questions :/

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    (Original post by creativebuzz)
    I'm really stuck on these questions :/

    d through f are just inverse chain rule

    g and h are too if you know how to work with odd powers of sin or cos

    For I consider the fourth root
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    (Original post by TenOfThem)
    d through f are just inverse chain rule

    g and h are too if you know how to work with odd powers of sin or cos

    For I consider the fourth root
    Awesome, I'll give that a shot!

    If you don't mind me asking, how did you know what to you by looking at it? I mean how did you look at d-f and tell that you need to use the inverse chain rule?
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    (Original post by creativebuzz)
    Awesome, I'll give that a shot!

    If you don't mind me asking, how did you know what to you by looking at it? I mean how did you look at d-f and tell that you need to use the inverse chain rule?
    You have g(f(x)) multiplied by f'(x)


    The first one does need to be turned into Cos(2x)Sin^-2(2x) first
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    (Original post by TenOfThem)
    You have g(f(x)) multiplied by f'(x)


    The first one does need to be turned into Cos(2x)Sin^-2(2x) first
    What do you mean?
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    (Original post by creativebuzz)
    What do you mean?
    Have you used the chain rule and the inverse chain rule
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    (Original post by TenOfThem)


    The first one does need to be turned into Cos(2x)Sin^-2(2x) first
    Why do you need to do that if you can integrat cosec(ax+b)cot(ax+b) = -1/a x cosec(ax + b)
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    (Original post by creativebuzz)
    Why do you need to do that if you can integrat cosec(ax+b)cot(ax+b) = -1/a x cosec(ax + b)
    Yup .... I forget that one ... It is so easy to derive using inverse chain rule

    Formula sheet rules ok
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    (Original post by TenOfThem)
    Yup .... I forget that one ... It is so easy to derive using inverse chain rule

    Formula sheet rules ok
    Ah cool! So I got -1/2cosec(2x) but when I put the limits in I managed to get some long decimals in my answer, am I plugging it into the calculator wrong or?

    (In my calculator I did sin(1/2 x pi/3) x -1/2 etc..)
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    (Original post by creativebuzz)
    Ah cool! So I got -1/2cosec(2x) but when I put the limits in I managed to get some long decimals in my answer, am I plugging it into the calculator wrong or?

    (In my calculator I did sin(1/2 x pi/3) x -1/2 etc..)
    It should be sin(2 x pi/3) then do 1/that
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    (Original post by TenOfThem)
    It should be sin(2 x pi/3) then do 1/that
    For the second one I got ln(2x-1) but when I substitute the limits in I can't seem to get the right answer! Where am I going wrong?
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    (Original post by creativebuzz)
    For the second one I got ln(2x-1) but when I substitute the limits in I can't seem to get the right answer! Where am I going wrong?
    What do you get if you differentiate ln(2x-1)

    You should see your error
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    (Original post by creativebuzz)
    For the second one I got ln(2x-1) but when I substitute the limits in I can't seem to get the right answer! Where am I going wrong?
    the question is wrong unless I cannot see well

    the integral does not converge between these limits
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    (Original post by TenOfThem)
    What do you get if you differentiate ln(2x-1)

    You should see your error
    You would get 1/(2x-1) x 2 = 1/4x-2 ... oh

    so can you take a factor of 2 out

    2ln(2x-1)
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    (Original post by creativebuzz)
    You would get 1/(2x-1) x 2 = 1/4x-2 ... oh

    so can you take a factor of 2 out

    2ln(2x-1)
    read post 13!
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    (Original post by creativebuzz)
    You would get 1/(2x-1) x 2 = 1/4x-2 ... oh

    so can you take a factor of 2 out

    2ln(2x-1)
    Hmmmmmmmmm

    1/(2x-1) x 2 = 2/(2x-1)

    So you need to divide your answer by 2 not multiply it
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    (Original post by TeeEm)
    the question is wrong unless I cannot see well

    the integral does not converge between these limits
    This is irrelevant in this context (sadly)

    At A2 Maths there is no learning related to the validity of an integral .... This is just a question in an exercise designed to test the student's understanding of the inverse chain rule
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    (Original post by TenOfThem)
    This is irrelevant in this context (sadly)

    At A2 Maths there is no learning related to the validity of an integral .... This is just a question in an exercise designed to test the student's understanding of the inverse chain rule
    ???

    then what answer will you give when there is no answer for these limits...
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    (Original post by TenOfThem)
    Hmmmmmmmmm

    1/(2x-1) x 2 = 2/(2x-1)

    So you need to divide your answer by 2 not multiply it
    Oh okay! But when I plug the limits into my calculator I get 0 for 1 and an error for -1
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    (Original post by TeeEm)
    ???

    then what answer will you give when there is no answer for these limits...
    The answer I would expect would be 0.5(ln1 - ln3) = -0.5ln3
 
 
 
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