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I've tried substituting $u=\cosh(t)$ whence

$$\int\nolimits_{\cosh^{-1}(0)}^{\cosh^{-1}(3)} \sqrt{\cosh^2(t)-\cosh{t}} dt$$ becomes

$$ \tag{1} \int _0^3 \sqrt{\frac{u}{u+1}}du $$ since $\sinh(\cosh^{-1}(u))=\sqrt{\frac{u^2-1}{u+1}}$ according to Wolfram Alpha. Equation (1) doesn't look that hard, and Wolfram Alpa gave the answer of $\sqrt{3} - \frac{1}{2}\sinh^{-1}(\sqrt{3})$, but I'd be much obliged if someone can give me some pointers as to how to evaluate this integral analytically/manually. I have a feeling it might be another substitution.

Thanks in advance.

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To my knowledge, $\cosh$ is always greater or equal than zero, so $\cosh ^{-1} (-3)$ cannot be a real number. – Oliver Braun Oct 24 '11 at 6:54
To evaluate the integral (1), you can make the substitution $u = \frac{t^2}{1-t^2} \iff \frac{u}{u+1} = t^2$. – Srivatsan Oct 24 '11 at 8:29
@OliverBraun. Thanks for that. You are right, I have edited the question. Hope it is right now. – Samuel Tan Oct 24 '11 at 8:36
@SrivatsanNarayanan. Clever substitution. Thank you. However, that means the integral becomes $$\int^{\frac{\sqrt{3}}{2}}_0 t dt = \frac{3}{2}$$ Is this right? – Samuel Tan Oct 24 '11 at 8:44
@SrivatsanNarayanan. Silly me! That means the integral becomes $$2\int^{\frac{\sqrt{3}}{2}}_0 \frac{t^2}{{\left(1-t^2\right)}^2} dt $$ Now it is right? Problem with the LaTex. – Samuel Tan Oct 24 '11 at 8:53
up vote 5 down vote accepted

One method is to make the substitution $$ u = \frac{t^2}{1-t^2} \iff t = \sqrt{\frac{u}{u+1}}. $$ However, this approach needs a bit of a struggle to complete.

A more clever approach (inspired by the answer given by Wolfram|Alpha) is to substitute $u = \sinh^2 t$. Then $$ \sqrt{\frac{u}{u+1}} = \sqrt{\frac{\sinh^2 t}{\cosh^2 t}} = \frac{\sinh t}{\cosh t}, $$ and $$ du = 2 \sinh t \cosh t. $$ Therefore, the integrand becomes: $$ \int_0^{\sinh^{-1}(\sqrt{3})} 2 \sinh^2 t dt = \frac 12 \int_0^{\sinh^{-1}(\sqrt{3})} (e^t - e^{-t})^2 dt = \frac 12 \int_0^{\sinh^{-1}(\sqrt{3})} (e^{2t} + e^{-2t} - 2) dt, $$ which can be easily evaluated.

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Nice answer. +1. – mick Jan 18 '13 at 16:54

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