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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 $\sinh(\cosh^{-1}(u))=\sqrt{\dfrac{u^2-1}{u+1}}$ according to Wolfram Alpha. Equation (1) doesn't look that hard, and Wolfram Alpha 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.

  • 0
    Again, please don't do that.2011-10-25

1 Answers 1

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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.

  • 0
    Nice answer. +1.2013-01-18