Somewhat inspired by this question about antiderivatives, I started to check whether or not that function had an elementary antiderivative. Then, after checking with Maxima, it struck me that, by simplifying the $\sec(x)$ and $\tan(x)$ terms using Taylor series, I could effectively solve the antiderivative of $\int {\sec\left(x\right)\tan\left(x\right) \over 3x + 5}\,{\rm d}x$.

Solving this, at origin 0 and with depth 8, I get the following expression.

$$ -\frac{9646207\,{x}^{9}}{1181250000}+\frac{9646207\,{x}^{8}}{630000000}-\frac{14069\,{x}^{7}}{875000}+\frac{14069\,{x}^{6}}{450000}-\frac{179\,{x}^{5}}{6250}+\frac{179\,{x}^{4}}{3000}-\frac{{x}^{3}}{25}+\frac{{x}^{2}}{10} $$

However, and back when I had Calculus, I never remembered using a Taylor series in order to solve an antiderivative.

Besides the resulting antiderivative being an approximation that degrades the further away the function is from the Taylor origin (as a Taylor series has a sort of implied error), what other faults or errors might happen should one use this technique?

  • $\begingroup$ In a typical second-semester calculus course in the USA, integration using Taylor series appears at the end; it's sort of a pinnacle of the course, tying in integrals and infinite series. $\endgroup$ – Post No Bulls Dec 31 '13 at 6:42
  • $\begingroup$ Yeah, that was a fun one... I still haven't figured it out. Guess my next foray is going to be Taylor Series. $\endgroup$ – Chris Feb 4 '14 at 19:18

There is nothing wrong with using Taylor's series for anti-derivative. In fact, the very first breakthrough in the computation of $\pi$ came when Gregory used your idea to get the anti-derivative for $\arctan$. Gauss did the same for $\arccos$. You are in good company!

  • $\begingroup$ I have to say I have yet to find anything about Gauss and $arccos$. This being said, a man by the name of Gregory Chudnovsky, together with his brother, developed a fast algorithm to calculate $\pi$. I'll keep this for a day or two, if no one gives another answer, I'll accept this. $\endgroup$ – Doktoro Reichard Dec 28 '13 at 23:03
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    $\begingroup$ The original proof of Sophomore's dream also used the sam idea as you did. Please see en.wikipedia.org/wiki/Sophomore%27s_dream. Chodonvsky's method is a bit different. $\endgroup$ – user44197 Dec 29 '13 at 4:48

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