I tried to develop the solution of this challenging integration but I never made it. Perhaps you could.. Integrating this term looked simpler but solving it stormed my mind. I found it rather difficult to integrate $\sqrt{\text{sin}x}$.
When I solve it , an infinte sequence arrived and I got trapped.
 A: As somments say, the result is not very nice since it involves elliptic integrals $$\int_0^x \sqrt {\sin(t)}~dt=\sqrt{\frac{2}{\pi }} \Gamma \left(\frac{3}{4}\right)^2-2 E\left(\left.\frac{1}{4}
   (\pi -2 x)\right|2\right)$$ What the comments say also is that you could try to use Wolfram Alpha or another CAS for these difficult problems.
For sure, and I suppose that this is what you tried, you could expand the integrand as an infinite series built at $t=0$ and integrate. You have $$\sqrt {\sin(t)}=\sqrt{t}-\frac{t^{5/2}}{12}+\frac{t^{9/2}}{1440}-\frac{t^{13/2}}{24192}-\frac{67
   t^{17/2}}{29030400}-\frac{t^{21/2}}{5677056}+O\left(t^{25/2}\right)$$ So $$\int_0^x \sqrt {\sin(t)}~dt=\frac{2
   t^{3/2}}{3}-\frac{t^{7/2}}{42}+\frac{t^{11/2}}{7920}-\frac{t^{15/2}}{181440}-\frac
   {67 t^{19/2}}{275788800}-\frac{t^{23/2}}{65286144}+O\left(t^{27/2}\right)$$ This leads to a very good approximation of the exact solution at least for $0\leq x\leq \pi$.
A: The fact that the antiderivatives of $\sin^ax$ and $\cos^ax$ cannot be expressed in terms of 
elementary functions for non-integer values of a, can ultimately be proven using either 
Liouville's theorem or the Risch algorithm. As already mentioned above, one needs special 
functions, such as elliptic integrals, in order to do that. However, for the definite integral, 
see Wallis' integrals. We have $\displaystyle\int_0^\tfrac\pi2\sin^ax~dx$ $=\dfrac{\dfrac\pi a}{B\bigg(\dfrac a2,\dfrac12\bigg)}$ , where $B(a,b)=\dfrac{\Gamma(a)\cdot\Gamma(b)}{\Gamma(a+b)}$ 
is the beta function, which extends the notion of binomial coefficients, and $\Gamma(n+1)=$
$=n\cdot\Gamma(n)$ is the $\Gamma$ function, which extends the factorial. In our case, for $a=\dfrac12$ , taking 
into consideration the fact that $\Gamma\bigg(\dfrac12\bigg)=\sqrt\pi$ in conjunction with Euler's reflection formula, 
the $($definite$)$ integral becomes $\dfrac{2\pi\sqrt{2\pi}}{\Gamma^2\bigg(\dfrac14\bigg)}~=~\sqrt{\dfrac2\pi}\cdot\Gamma^2\bigg(\dfrac34\bigg)$. Hope this helps.
