I came over the following integral $$ \int_0^\pi \sin(x^3) \mathrm{d}x $$ when a friend of mine tried to approximate it. The most obvious way is to use taylors formula, and then turn the integral into a sum. Eg $$ \int_0^\pi \sin(x^3) \mathrm{d}x \approx \sum_{k=0}^N \frac{1}{2} \frac{(-1)^k \pi^{3k+2}}{(3k+2)(2k+1)!} $$ The problem is the nominator increases much more rapidly than the denominator for the first 30 terms or so. Much faster than what a standard calculator can deal with. This can be seen here https://i.stack.imgur.com/O8aX0.jpg
By using the midpoint rule instead, the sum turns into $$ \int_0^\pi \sin(x^3) \mathrm{d}x \approx \frac{\pi}{2}\sum_{k=0}^N f\left( \frac{\pi}{2} \frac{2k-1}{N}\right) $$ Which converges to two decimal places in $36$ iterations, however to obtain a higher accuracy the convergence is slower than the taylor series.
As a final note I also tried using the substitution $u \mapsto u^3$, but the series expansion did not improve.
My question is as follows: What is the least terms needed to approximate the integral to $3$ digits accuracy under the restrictions of using a standard pocket calculator?
To clearify the calculator does not have more than 8 digits accuracy, and can use sine and cosine. Oh, I also tried Simpsons that did not improve the convergence. Could Romberg, or Gaussian lead to faster convergence?