so it goes
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Report Thread starter 6 years ago
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I'm asked to show that:

\int{x arcosh x}dx = \frac{1}{4}(2x^2 -1)arcosh x - \frac{1}{4}x\sqrt{x^2 -1} + C

If I integrate by parts:

let u = arcosh x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{\sqrt{x^2 -1}}

and let \dfrac{dv}{dx} = x \Rightarrow v = \frac{1}{2}x^2

using the formula for integration by parts and rearranging gives:

I = \frac{1}{2}x^2 arcosh x - \frac{1}{2}\int{\sqrt{x^2 - 1} + \dfrac{1}{\sqrt{x^2 -1}}}dx

If I use the substitution:

let x = \cosh u

Using this substitution and rearranging gives:

I = \frac{1}{2}x^2 arcosh x - \frac{1}{2}\int{\cosh^2 u} du

\Rightarrow I = \frac{1}{2}x^2 arcosh x - \frac{1}{4}[\sinh u \cosh u + u] + C

I want to find this in terms of x. To eliminate u, I will find \sinh u in terms of x.

I know that \cosh u = x

\Rightarrow \cosh^2 u = x^2

\Rightarrow \sinh^2 u + 1 = x^2

\Rightarrow \sinh^2 u = x^2 - 1

\Rightarrow \sinh u = \pm \sqrt{x^2 - 1}

I can therefore say:

I = \frac{1}{2}x^2 arcosh x - \frac{1}{4}[\pm x\sqrt{x^2 - 1} + arcosh x] + C

\Rightarrow I = \frac{1}{4}(2x^2 - 1)arcosh x \pm \frac{1}{4}x\sqrt{x^2 - 1} + C

Have I misunderstood something about \sinh u?

Could someone please explain why I must ignore the negative option when square rooting \sinh^2 u?

Thank you.
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electriic_ink
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Not done FP3 in a while but arcosh x seems to me to a lot like sqrt(x), in that it's a poorly defined function without some restriction on the range, like arcosh x > 0. So if u=cosh x, we must assume x>0 and so that sinh x > 0.

Of course, we could instead assume that arcosh x <0. But there must be the convention that arcosh x > 0 listed in the textbook/some restriction given in the question that you are expected to follow.
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Farhan.Hanif93
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The subtlety is in the square rooting process early on. Notice that, when you make the substitution originally, your right hand integral boils down to:

\displaystyle\int \left(\sqrt{\sinh^2 u} + \dfrac{1}{\sqrt{\sinh ^2u}}\right)\sinh u du

But, in general, \sqrt{x^2} = |x|, so \sqrt{\sinh ^2 u} = \sinh u \iff \sinh u\geq 0 \iff u\geq 0.

Hence, implicitly in your working, you have assumed that u\geq 0 when working towards \displaystyle\int \cosh ^2 u du; and therefore, you must carry forward this convention when inverting the substitution - i.e. that \sinh u \geq 0 so that x=\cosh u \implies \sinh u = +\sqrt{x^2-1}.


[Therefore, your result only actually holds for x\geq 1 (as this is where x=\cosh u makes sense as a substitution) - However, this is fine in this case, since the original integral \displaystyle\int x\text{arccosh} \ x dx only makes sense where \text{arccosh} x is defined, which is conventionally for x&gt;1 in FP3 (as electric_ink suggested)]
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so it goes
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Report Thread starter 6 years ago
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(Original post by electriic_ink)
Not done FP3 in a while but arcosh x seems to me to a lot like sqrt(x), in that it's a poorly defined function without some restriction on the range, like arcosh x > 0. So if u=cosh x, we must assume x>0 and so that sinh x > 0.

Of course, we could instead assume that arcosh x <0. But there must be the convention that arcosh x > 0 listed in the textbook/some restriction given in the question that you are expected to follow.

(Original post by Farhan.Hanif93)
The subtlety is in the square rooting process early on. Notice that, when you make the substitution originally, your right hand integral boils down to:

\displaystyle\int \left(\sqrt{\sinh^2 u} + \dfrac{1}{\sqrt{\sinh ^2u}}\right)\sinh u du

But, in general, \sqrt{x^2} = |x|, so \sqrt{\sinh ^2 u} = \sinh u \iff \sinh u\geq 0 \iff u\geq 0.

Hence, implicitly in your working, you have assumed that u\geq 0 when working towards \displaystyle\int \cosh ^2 u du; and therefore, you must carry forward this convention when inverting the substitution - i.e. that \sinh u \geq 0 so that x=\cosh u \implies \sinh u = +\sqrt{x^2-1}.


[Therefore, your result only actually holds for x\geq 1 (as this is where x=\cosh u makes sense as a substitution) - However, this is fine in this case, since the original integral \displaystyle\int x\text{arccosh} \ x dx only makes sense where \text{arccosh} x is defined, which is conventionally for x&gt;1 in FP3 (as electric_ink suggested)]
Ahh, of course! This all makes sense now.

Thank you both

Sorry Farhan.Hanif93 but TSR isn't letting me up-vote your comment or give you rep - the sentiment is there though
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