Specifically, this is the integral I am talking about:
$$ \int_0^x \sin(t^6) \,dt $$
To integrate this, I used the exponential representation of trig functions:
$$ - \int \frac{i(e^{it^6} - e^{-it^6})}{2} \ dt $$
After I applied linearity and u-subsitution,
$$ \frac{i}{2} \int e^{-it^6} \ dt - \frac{i}{2} \int e^{it^6} \ dt $$ $$ u = t\sqrt[\leftroot{-2}\uproot{2}6]{-i} \,, \frac{du}{dt} = \sqrt[\leftroot{-2}\uproot{2}6]{-i} $$
I arrive here with this indefinite integral:
$$ \int e^{u^6} du $$
which is apparently a special integral that represents the incomplete gamma function.
$$ -\dfrac{\operatorname{\Gamma}\left(\frac{1}{6},u^6\right)}{6} $$
Anyhow, following the steps through and simplifing, at the end I got this answer for the indefinite integral portion:
$$-\dfrac{\left(-\mathrm{i}\right)^\frac{5}{6}\left(\left(-1\right)^\frac{5}{6}\operatorname{\Gamma}\left(\frac{1}{6},\mathrm{i}t^6\right)+\operatorname{\Gamma}\left(\frac{1}{6},-\mathrm{i}t^6\right)\right)}{12}+C$$
Solving for the definite integral from bounds $0$ to $x$, I get this as the final answer:
$$-\dfrac{\left(\operatorname{\Gamma}\left(\frac{1}{6},\mathrm{i}x^6\right)+\operatorname{\Gamma}\left(\frac{1}{6},-\mathrm{i}x^6\right)-2\operatorname{\Gamma}\left(\frac{1}{6}\right)\right)\sin\left(\frac{{\pi}}{12}\right)+\left(\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{6},\mathrm{i}x^6\right)-\mathrm{i}\operatorname{\Gamma}\left(\frac{1}{6},-\mathrm{i}x^6\right)\right)\cos\left(\frac{{\pi}}{12}\right)}{12}$$
Now, I am not familiar with the gamma function, however I am familiar in plain old series. Is it possible to represent the above solution just using fractions and factorials?
Someone told me it's possible but they weren't sure how either.
$$\int_0^x\sum_{k=0}^\infty(-1)^k\frac{(t^6)^{2k+1}}{(2k+1)!}dt=\sum_{k=0}^\infty(-1)^k\frac{x^{12k+7}}{(12k+7)(2k+1)!}.$$