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    • PS Reviewer
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    \int cosh^3 x

    The problem is i'm having difficulty proving the following identity:

    cosh^3 x = \dfrac{1}{4}cosh 3x + \dfrac{3}{4}cosh x

    If that's even right.

    I kind of get stuck at cosh x(1+sinh^2 x)

    Thanks

    Sohan
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    Try expanding \cosh 3x.
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    Why dont you say cosh^2(x) = 1 + sinh^2(x)

    cosh^3(x) = cosh(x) + cosh(x)sinh^2(x)

    Then cosh(x) integrates to give sinh(x)
    and cosh(x)sinh^2(x)... well use the substitution t = sinh(x)

    And then you don't even need to bother proving that identity
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    (Original post by Glutamic Acid)
    Try expanding \cosh 3x.
    Yeah but that's practically cheating. In the exam, I won't have the answer to work backwards from.
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    Just see \displaystyle \int ch ^3 = \int \h(1 + sh^2) = \int ch + sh ^2 ch. Then first term is easier, and second term is a function to a power, next to its derivative (with some coefficient)
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    (Original post by tazarooni89)
    Why dont you say cosh^2(x) = 1 + sinh^2(x)

    cosh^3(x) = cosh(x) + cosh(x)sinh^2(x)

    Then cosh(x) integrates to give sinh(x)
    and cosh(x)sinh^2(x)... well use the substitution t = sinh(x)

    And then you don't even need to bother proving that identity
    Yeah I neevr thought to do it like that :banghead:

    Thanks
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    (Original post by sohanshah)
    Yeah but that's practically cheating. In the exam, I won't have the answer to work backwards from.
    Oh, I see, I thought that result was in the question. Use Kolya's method.
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    Any idea as to how to \displaystyle \int cosech\frac{x}{3} dx
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    Well... first you need to know how to integrate cosech(x) and then it's easy

    If you treat cosech(x) as a fraction, [cosech(x)]/1

    then multiply the top and bottom of that fraction by coth(x) - cosech(x), you should notice that the top of the fraction is the differential of the bottom.

    Does that help?
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    Use the t-substitution. If you aren't sure how to formulate it for hyperbolic, just write \sinh u = -i\sin (iu), take the factor of i outside of the integral sign, integrate using the t-substitution, and then turn it back into hyperbolic functions afterwards.
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    Multiply top and bottom by \sinh \frac{x}{3}
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    (Original post by tazarooni89)
    then multiply the top and bottom of that fraction by coth(x) - cosech(x), you should notice that the top of the fraction is the differential of the bottom.
    Divine inspiration, eh? :wink2:
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    Half tanh substitution:

    t=\tanh(\frac{\theta}{2})

    d\theta=\frac{2dt}{1-t^2}

    \mathrm{cosech}(\theta)=\frac{1-t^2}{2t}
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    (Original post by Kolya)
    Divine inspiration, eh? :wink2:
    lol, no that's just how my FP3 teacher taught it :rolleyes:

    If I were integrating cosech(x) during my university course, I'd probably use the substitution t = sinh x, which makes it work out quite nicely, although I don't think an FP3 question requires you to be super-rigorous about it.
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    (Original post by Mr M)
    Half tanh substitution:

    t=\tanh(\frac{\theta}{2})

    d\theta=\frac{2dt}{1-t^2}

    \mathrm{cosech}(\theta)=\frac{1-t^2}{2t}
    How did you get from step 2 to 3??

    I'm basically self-teaching this so don't think i'm stupid. Even though I am!
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    (Original post by sohanshah)
    How did you get from step 2 to 3??

    I'm basically self-teaching this so don't think i'm stupid. Even though I am!
    Are you familar with half tangent substutions? If not, private message me with your email address and I will send you a Powerpoint about it.

    From the half tangent subs, you just apply Osbourne's rule (so the signs change for t^2).
 
 
 
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