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    Integral of sqrt of (1+t^2), using the substitution t=sinhx ?
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    (Original post by RockConcert)
    Integral of sqrt of (1+t^2), using the substitution t=sinhx ?
    just think of it as a normal C4 t=sinx sub except be careful with the identities (you know how the hyperbolic identities differ slightly from the trig ones)
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    (Original post by DylanJ42)
    just think of it as a normal C4 t=sinx sub except be careful with the identities (you know how the hyperbolic identities differ slightly from the trig ones)
    I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )
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    (Original post by RockConcert)
    I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )
    maybe im wrong, let me try it, 2 secs
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    integral becomes 1 + sinh^2(x)/cosh(x)

    change 1 + sinh^2(x)
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    (Original post by RockConcert)
    I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )
    Show us your working. We can't help until you do.

    Oh, and moved to maths.
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    (Original post by RockConcert)
    I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )
    okay got it, can you post some workings?
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    (Original post by RockConcert)
    I keep trying with that approach but my answer is far too simple.. ( arsinht instead of 1/2(t(sqrt of (1+t^2))+arsinht) )
    You put the differential in 'upside down'
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    (Original post by RockConcert)
    Attachment 511355
    As above, it should be \int \cosh x \sqrt{1 + \sinh^2 x} \, \mathrm{d}x.
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    Dt=cosh(X)dx is what you needed


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    (Original post by RockConcert)
    You got confused with your substitution. We want :

     \mathrm{d} t = \mathrm{d}x \cosh{x}

    And the answer will follow using the hyperbolic equivalent of  \sin{2\theta} = 2\cos{\theta} \sin{\theta}

    You should have recognised your mistake right away as  \sqrt{1+t^2} \neq \dfrac{1}{\sqrt{1+t^2}} .
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    (Original post by Louisb19)
    And the answer will follow using the hyperbolic equivalent of  \sin{2\theta} = 2\cos{\theta} \sin{\theta}
    Surely you mean \cos 2\theta = 2\cos^2 \theta - 1?
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    (Original post by Zacken)
    Surely you mean \cos 2\theta = 2\cos^2 \theta - 1?
    i think he means after the integration has been done, using that identity to get the answer into the correct form
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     \displaystyle \cosh 2x=2\cosh^2 x -1 .
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    (Original post by DylanJ42)
    i think he means after the integration has been done, using that identity to get the answer into the correct form
    Ah, okay. :facepalm:
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    (Original post by DylanJ42)
    i think he means after the integration has been done, using that identity to get the answer into the correct form
    true dat PRSOM
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    (Original post by Louisb19)
    You got confused with your substitution. We want :

     \mathrm{d} t = \mathrm{d}x \cosh{x}

    And the answer will follow using the hyperbolic equivalent of  \sin{2\theta} = 2\cos{\theta} \sin{\theta}

    You should have recognised your mistake right away as  \sqrt{1+t^2} \neq \dfrac{1}{\sqrt{1+t^2}} .
    Ah thank you so much, that identity (sin2x) was the key! Man I'm not going to forget that identity again..
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    (Original post by RockConcert)
    Ah thank you so much, that identity (sin2x) was the key! Man I'm not going to forget that identity again..
    remember it with the h on the end too. cosh
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    (Original post by EricPiphany)
    remember it with the h on the end too. cosh

    Don't worry that was a typo, I did use sinh2x = 2sinhxcoshx in my solution. But thank you for your concern!
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    (Original post by RockConcert)
    Don't worry that was a typo, I did use sinh2x = 2sinhxcoshx in my solution. But thank you for your concern!
    You're welcome xxx
 
 
 
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