Sketch the integrand!

First, one can use the theorem of change in variable for $u=x-\pi \rightarrow du=dx$ , and interval of integration becomes $[\frac{\pi}{2}-\pi, \frac{3\pi}{2}-\pi] \rightarrow [\frac{-\pi}{2}, \frac{\pi}{2}]$ . Then:

$\displaystyle I=\frac{1}{\pi}.\sqrt{\frac{2}{\pi}}.\int_{-\pi/2}^{\pi/2}(u^2)^{\frac{1}{4}}.sin(u+\pi)+\sqrt{\frac{\pi}{2}} \ du$

But $sin(u+\pi)=-sin(u)$ and $-(u^2)^{\frac{1}{4}}.sin(u)$ is an odd function integrated over a symmetrical interval (i. e. null total area). Splitting integrals:

$\displaystyle I=\frac{1}{\pi}.\sqrt{\frac{2}{\pi}}.\int_{-\pi/2}^{\pi/2}-(u^2)^{\frac{1}{4}}.sin(u) \ du+\frac{1}{\pi}.\sqrt{\frac{2}{\pi}}.\int_{-\pi/2}^{\pi/2}\sqrt{\frac{\pi}{2}} \ du$

$\displaystyle I=\frac{1}{\pi}.\sqrt{\frac{2}{\pi}}.\int_{-\pi/2}^{\pi/2}\sqrt{\frac{\pi}{2}} \ du=\frac{1}{\pi}.\int_{-\pi/2}^{\pi/2} 1 \ du=\boxed{1}$