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My friend took his Calculus $2/3$ test yesterday.

One of the questions he had trouble with was this integral:

$$\int \sqrt{t^4-t^2 + 1}dt$$

My attempt

It seems rather clear that the only approach was trigonometric substitution.

First, completing the square:

$$\int \sqrt{t^4-t^2 + 1}dt = \int\sqrt{t^4-t^2 + \frac{1}{4} + \frac{3}{4}}dt = \int \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}dt$$

Next, I let $$\sec \theta = \frac{\sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}}{\frac{\sqrt{3}}{2}}$$ $$\frac{\sqrt{3}}{2}\sec \theta = \sqrt{\left(t^2 - \frac{1}{2}\right) + \frac{3}{4}}$$ $$\tan \theta = \frac{t^2 - \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{2t^2 - 1}{\sqrt{3}}$$ $$\sec^2 \theta d\theta = \frac{4}{\sqrt{3}}tdt,\ dt = \frac{\sqrt{3}\sec^2 \theta}{4t} d\theta$$ $$t = \sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}$$


Substituting this all in:

$$\int \left(\frac{\sqrt{3}}{2} \sec \theta\right) \frac{\sqrt{3}\sec^2 \theta}{4\sqrt{\frac{1}{2}\left(\sqrt{3}\tan \theta + 1\right)}} d\theta$$

How would I approach this from here?

I'm thinking of using u-substitution but I'm sure that it would bring me back to where I started, meaning I would have to use trig. substitution again.

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  • $\begingroup$ This looked pretty unmanageable, and WolframAlpha gives an answer involving elliptic integral functions, so presumably there is no elementary closed form. Perhaps the problem was (intended to be) $\int \sqrt{t^4 - 2 t^2 + 1} \,dt$ or something similar? $\endgroup$ – Travis Nov 13 '14 at 11:43
  • $\begingroup$ @Travis If that is the case, then maybe there is no closed form. It seemed to me that there would be. $\endgroup$ – Varun Iyer Nov 13 '14 at 11:45
  • $\begingroup$ I know that you can use euler substitution when you got $\sqrt(at^2+bt+c)$, maybe there is some way you can use euler substitution on this? Also I think it is going to be $(t^2-\frac{1}{2})^2$ instead of $(t^2-\frac{1}{2})$ $\endgroup$ – harbor Nov 13 '14 at 12:41
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    $\begingroup$ You forgot to raise your quantity $$\left(t^2-\frac{1}{2}\right)$$ to a power of two. $\endgroup$ – graydad Nov 13 '14 at 14:34
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The indefinite integral cannot be expressed in terms of elementary functions. Evaluating it requires knowledge of elliptic integrals. However, it is interesting to note that its definite counterpart, when evaluated over the entire real line, and subtracted from its quadratic or parabolic asymptote, yields

$$\int_{-\infty}^\infty\bigg[\sqrt{t^4-t^2+1}-\bigg(t^2-\dfrac12\bigg)\bigg]dt~=~\dfrac23K\bigg(\dfrac34\bigg)+\dfrac43E\bigg(\dfrac34\bigg),$$

which is nothing else than the volume of the oloid, whose decimal expansion can be found on OEIS

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