Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$ It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is 
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are your ideas on it? Differentiation under the integral sign? Other ways?
Mathematica 9 says that 
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx=\frac{2}{3} \, _5F_4\left(\frac{1}{2},\frac{2}{3},1,1,\frac{4}{3};\frac{5}{6},\frac{7}{6},\frac{3}{2},\frac{3}{2};-1\right).$$
 A: Just exploiting the arctangent Taylor series,
$$\begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3 x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\int_{0}^{\frac{\pi}{2}}\sin^{6m+3}(x)\,dx\\ &=& \frac{3\sqrt{\pi}}{4}\sum_{m\geq 0}\frac{(-1)^m\,\Gamma\left(3m+2\right)}{(3m+\frac{3}{2})\Gamma\left(3m+\frac{5}{2}\right)}\end{eqnarray*} $$
so, at least in principle, we may compute the RHS by applying a discrete Fourier transform to the power series:
$$ \sum_{m\geq 0}\frac{x^m\, \Gamma(m+2)}{\left(m+\frac{3}{2}\right)\Gamma\left(m+\frac{5}{2}\right)}$$
that is just a $\phantom{}_{3} F_2$ hypergeometric function, namely $\frac{8}{9\sqrt{\pi}}\;\phantom{}_3 F_2\!\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x\right).$
A: This will be too long for a comment, but I am not able to give the solution, just a short form of the answer. I'll tell you how to get there (with your favorite CAS).
First, note that (using $x\mapsto \pi/2-x$)
$$
I=\int_0^{\pi/2}\arctan(\sin^3x)\,dx=\int_0^{\pi/2}\arctan(\cos^3x)\,dx
$$
Thus, integrating by parts,
$$
\begin{aligned}
I&=\frac{1}{2}\int_0^{\pi/2} 1\bigl(\arctan(\sin^3x)+\arctan(\cos^3x)\bigr)\,dx\\
&=\frac{1}{2}\Bigl[x\bigl(\arctan(\sin^3x)+\arctan(\cos^3x)\bigr)\Bigr]_0^{\pi/2}\\
&\quad -\frac{1}{2}\int_0^{\pi/2}x\Bigl(\frac{3\sin^2x\cos x}{1+\sin^6x}-\frac{3\cos^2x\sin x}{1+\cos^6 x}\Bigr)\,dx
\end{aligned}
$$
The out-integrated part is $\pi^2/8$ or something like that, and the other part does the CAS take care of, and the result is
$$
\begin{aligned}
I&=\frac{1}{2}\text{Li}_2(3-2\sqrt{2})-2\text{Li}_2(\sqrt{2}-1)-\text{Li}_2\bigl(-(\sqrt{3}-1)(1+i)/2\bigr)-\text{Li}_2\bigl(-(\sqrt{3}-1)(1-i)/2\bigr)\\
&\qquad+\text{Li}_2\bigl((\sqrt{3}-1)(1-i)/2\bigr)+\text{Li}_2\bigl((\sqrt{3}-1)(1+i)/2\bigr)\\
\end{aligned}
$$
Numerically, this evaluates to approximately $0.571665$.
I'm not too familiar with polylogarithms, so maybe someone else would like ty try to simplify this expression further.
