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Integral of 4Tan3x, and AQA Formula booklet

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    Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.

    Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.

    so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ?

    I am sorry if this sounds confusing btw!
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    EDIT: ignore this sorry i found further down that the integral of tankx is something else, sorry!
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    (Original post by raveen789)
    Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.

    Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.

    so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ?

    I am sorry if this sounds confusing btw!
     \displaystyle \int 4tan3x \ dx = \frac43 ln|sec3x| + C

    You are doing correct.
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    (Original post by raheem94)
     \displaystyle \int 4tan3x \ dx = \frac43 ln|sec3x| + C

    You are doing correct.
    oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!
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    (Original post by raveen789)
    oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!
    It would probably be  \displaystyle \int tankx \ dx = - \frac1k ln|coskx| + C
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    (Original post by raheem94)
    It would probably be  \displaystyle \int tankx = - \frac1k ln|coskx| + C
    Why the minus sign?
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    (Original post by ghostwalker)
    Why the minus sign?
     \displaystyle \int tankx \ dx = - \frac1k ln|coskx| + C

     \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k ln|seckx|

    Please correct me if i am wrong.
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    (Original post by raheem94)
     \displaystyle \int tankx = - \frac1k ln|coskx| + C

     \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k|seckx|

    Please correct me if i am wrong.
    Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).

    Ignore: Irrelevant.
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    (Original post by ghostwalker)
    Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).
    Maybe I'm missing something but I don't see the relevance?
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    (Original post by ghostwalker)
    Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).
    These aren't hyperbolic functions.

    It is a 'k' not a 'h'.

    Where is the typo, i don't see it?
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    (Original post by hassi94)
    Maybe I'm missing something but I don't see the relevance?
    Is the formula i have written correct?

    Sorry for missing the 'dx', i forgot to write it.
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    (Original post by raheem94)
     \displaystyle \int tankx = - \frac1k ln|coskx| + C

     \displaystyle - \frac1k ln|coskx| = \frac1k ln|coskx|^{-1} = \frac1k|seckx|

    Please correct me if i am wrong.
    (Original post by hassi94)
    Maybe I'm missing something but I don't see the relevance?
    I'm being a moron!

    Reading k as h, and thinking it's hyperbolics. Duh!

    Ignore me: I'll just go and shoot myself.
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    (Original post by raheem94)
    These aren't hyperbolic functions.

    It is a 'k' not a 'h'.

    Where is the typo, i don't see it?
    The typo was you missing 'ln' on the last bit of the second line

    Otherwise I think it's correct.
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    (Original post by ghostwalker)
    I'm being a moron!

    Reading k as h, and thinking it's hyperbolics. Duh!

    Ignore me.
    Aha okay that makes sense now, don't worry about it
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    (Original post by ghostwalker)
    I'm being a moron!

    ((((hugs)))) for ghostwalker

    we have all been there
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    (Original post by ghostwalker)
    I'm being a moron!

    Reading k as h, and thinking it's hyperbolics. Duh!

    Ignore me.
    Never knew you also make mistakes!

    (Original post by hassi94)
    The typo was you missing 'ln' on the last bit of the second line

    Otherwise I think it's correct.
    Thanks i will edit my post, its too difficult to spot your own mistakes.
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    Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/
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    (Original post by raveen789)
    Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/
    Yeah it seems fairly hit and miss. Just remember you must as it's 'the reverse chain rule'.

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