# Evaluation of the integral $\int \sqrt{t^4-t^2 + 1}\,dt$

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

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

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.

• 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? – Travis Nov 13 '14 at 11:43
• @Travis If that is the case, then maybe there is no closed form. It seemed to me that there would be. – Varun Iyer Nov 13 '14 at 11:45
• 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 $(t^2-\frac{1}{2})$ – harbor Nov 13 '14 at 12:41
• You forgot to raise your quantity $$\left(t^2-\frac{1}{2}\right)$$ to a power of two. – graydad Nov 13 '14 at 14:34

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