Evaluate $\int_0^{\pi/2}\frac{1-\sqrt[18]{\cos u}}{1-\cos u}du$, in terms of particular values of special functions and constants I wondered about this question when I've considered similar integrals with different integrand as an analogous of a formula for harmonic number. 
As example that I know that can be calculated using Wolfram Alpha (it calculate the closed-form for the definite integral, and also the indefinite integral), I would like to know how get the result.

Question. Can you calculate, justifying details or hints, in closed-form $$\int_0^{\frac{\pi}{2}}\frac{1-\sqrt[18]{\cos u}}{1-\cos u}du\,?$$
  Thanks in advance.

I've choosen the root $18$ since I don't know if a more simple example was in the literature.
 A: $$\int_{0}^{\pi/2}\frac{1-\sqrt[18]{\cos x}}{1-\cos x}\,dx =\int_{0}^{\pi/4}\frac{1-\sqrt[18]{2\cos^2 x-1}}{\sin^2 x}\,dx=\int_{0}^{1}\frac{1-\sqrt[18]{\frac{1-t^2}{1+t^2}}}{t^2}\,dt $$
equals
$$ \frac{1}{2}\int_{0}^{1}\frac{1-\sqrt[18]{\frac{1-t}{1+t}}}{t\sqrt{t}}\,dt=\int_{0}^{1}\frac{1-\sqrt[18]{u}}{(1-u)\sqrt{1-u^2}}\,du$$
which can be managed through integration by parts and Euler's Beta function. The final outcome is
$$ -1+\sqrt{\pi}\left[\frac{\Gamma\left(\frac{1}{36}\right)}{36\,\Gamma\left(\frac{19}{36}\right)}+\frac{\Gamma\left(\frac{19}{36}\right)}{\Gamma\left(\frac{1}{36}\right)}\right]\approx \frac{499}{4088}.$$
The approach is the same if $18$ is replaced by any other $m>0$: you'll get an expression involving $\Gamma\left(\frac{1}{2m}\right)$ and $\Gamma\left(\frac{1}{2}+\frac{1}{2m}\right)$.
A: This answer is similar to Jack D'Aurizio's answer, but I thought a bit more care might be needed near $u=1$.
$$
\begin{align}
\int_0^{\frac\pi2}\frac{1-\sqrt[n]{\cos(u)}}{1-\cos(u)}\,\mathrm{d}u
&=\int_0^1\frac{1-\sqrt[n]{u}}{1-u}\frac{\mathrm{d}u}{\sqrt{1-u^2}}\tag1\\
&=\int_0^1\frac{\left(1-\sqrt[n]{u}\right)(1+u)}{\sqrt{1-u^2}^3}\,\mathrm{d}u\tag2\\
&=\frac12\int_0^1\frac{\left(1-\sqrt[2n]{u}\right)\left(1+\frac1{\sqrt{u}}\right)}{\sqrt{1-u}^3}\,\mathrm{d}u\tag3\\
&=\frac12\int_0^1\left(1-u^{\frac1{2n}}\right)(1-u)^{-\frac32}\,\mathrm{d}u\\
&+\frac12\int_0^1\left(u^{-\frac12}-u^{-\frac{n-1}{2n}}\right)(1-u)^{-\frac32}\,\mathrm{d}u\tag4\\
&=\frac12\frac{\Gamma(1)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac12\right)}-\frac12\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac12+\frac1{2n}\right)}\\
&+\frac12\frac{\Gamma\left(\frac12\right)\Gamma\left(-\frac12\right)}{\Gamma(0)}-\frac12\frac{\Gamma\left(\frac12+\frac1{2n}\right)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac1{2n}\right)}\tag5\\
&=-1+\sqrt\pi\left(\frac{\Gamma\left(\frac{2n+1}{2n}\right)}{\Gamma\left(\frac{n+1}{2n}\right)}+\frac{\Gamma\left(\frac{n+1}{2n}\right)}{\Gamma\left(\frac1{2n}\right)}\right)\tag6\\
&=-1+\sqrt\pi\left(\frac{\Gamma\left(\frac{1}{2n}\right)}{2n\,\Gamma\left(\frac{n+1}{2n}\right)}+\frac{\Gamma\left(\frac{n+1}{2n}\right)}{\Gamma\left(\frac1{2n}\right)}\right)\tag7
\end{align}
$$
Explanation:
$(1)$: substitute $u\mapsto\cos^{-1}(u)$
$(2)$: multiply numerator and denominator by $1+u$
$(3)$: substitute $u\mapsto\sqrt{u}$
$(4)$: distribute $1$ and $\frac1{\sqrt{u}}$ over two integrals
$(5)$: apply $(8)$
$(6)$: collect and simplify using $\Gamma\!\left(\frac12\right)=\sqrt\pi$ and $\Gamma\!\left(-\frac12\right)=-2\sqrt\pi$
$(7)$: $\Gamma(1+x)=x\,\Gamma(x)$

Justification for the non-standard Beta function limits
Let $a,b\gt0$ and $c\gt-1$. Then we can write $1=t+(1-t)$ and integrate by parts to get
$$
\begin{align}
&\int_0^1\left(t^{a-1}-t^{b-1}\right)(1-t)^{c-1}\,\mathrm{d}t\\
&=\int_0^1\left(t^a-t^b\right)(1-t)^{c-1}\,\mathrm{d}t
+\int_0^1\left(t^{a-1}-t^{b-1}\right)(1-t)^c\,\mathrm{d}t\\
&=\frac1c\int_0^1\left((a+c)t^{a-1}-(b+c)t^{b-1}\right)(1-t)^c\,\mathrm{d}t\\[3pt]
&=\frac{\Gamma(a)\,\Gamma(c)}{\Gamma(a+c)}-\frac{\Gamma(b)\,\Gamma(c)}{\Gamma(b+c)}\tag8
\end{align}
$$
which can also be gotten by analytic continuation from the formula for $c\gt0$.
