In the book, "Pi and the AGM" by Borwein and Borwein, it is mentioned that Gauss computed the following integral to the eleventh decimal palce.


$\int\_0^1 \frac{1}{\sqrt{1-x^4}}dx$


How did he do it? Personnally, I looked at a Taylor expansion of


$\frac{1}{\sqrt{1-x}}$


Where I substituted $t^4$ for $x$, and integrated term by term, but this gives a series that converges really slowly. Is there an obvious transform to make this computation faster?



Numerically computing $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$
A good place to look is pages 405 and 413 of the *Nachlass* section of Gauss's *Werke III*, which can be found online through Google Books. On page 405, he gives the following formula for "$\text{arc sin lemn }x$":


$$\text{arc sin lemn }x= x+{1\over2}\cdot{1\over5}x^5 + {1\cdot3\over2\cdot4}{1\over9}x^9+{1\cdot3\cdot5\over2\cdot4\cdot6}{1\over13}x^{13}+{1\cdot3\cdot5\cdot7\over2\cdot4\cdot6\cdot8}{1\over17}x^{17}+\cdots$$


One page 413, he computes the value of $\int\_0^1{dx\over\sqrt{1-x^4}}$, presumably using the expansion above, but explicitly citing the formula


$$\text{ arc sin lemn }{7\over23}+2\text{arc sin lemn }{1\over2}$$


obtaining


$$1.3110287771\quad460599052\quad320.7$$


He also notes there that Stirling had obtained the value $1.3110287771\ 4605987$. This is in reference to the calculations on pages 57-58 of Stirling's *Methodus differentialis* from 1730, which can also be found through Google Books. It might be worth noting that even Gauss was slightly off in the last couple of decimal places. A more accurate value, which I took from [here](http://www2.mae.ufl.edu/~uhk/COMPLETE-ELLIPTIC-INTEGRALS.pdf), is


$$1.3110287771\quad460599052\quad324197949$$


