Put,

$t=tan(x/2), sin\,x=\frac{2t}{1+t^2}, cos\,t= \frac{1-t^2}{1+t^2}, dx= \frac{2\,dt}{1+t^2}$

When,

$x=0, t=0, x= \pi, t= \infty, $

$I= \int_{0}^{\infty} \frac{\sqrt{\frac{2t}{1+t^2}}}{(5+3(\frac{1-t^2}{1+t^2}))^{3/2}} \frac{2dt}{(1+t^2)}$

$= \int_{0}^{\infty} \frac{2\sqrt{2}\sqrt{t}}{(8+2t^2)^{3/2}}dt = \int_{0}^{\infty} \frac{\sqrt{t}}{(4+t^2)^{3/2}}dt$

Put,

$t^2= 4y, t=2\sqrt{y}, dt=\frac{dy}{\sqrt{y}}$

$I=\frac{1}{8} \int_{0}^{\infty} \frac{\sqrt{2}\,y^{1/4}}{(1+y)^{3/2}}\,\frac{dy}{\sqrt{y}}$

$=\frac{1}{4\sqrt{2}} \int_{0}^{\infty}\frac{y^{-1/4}}{(1+y)^{3/2}}dy= \frac{1}{4\sqrt{2}} \beta(\frac{3}{4}, \frac{3}{4}) $

$= \frac{1}{4\sqrt{2}} \frac{\sqrt{3/4}\sqrt{3/4}}{\sqrt{\frac{3}{4}+\frac{3}{4}}} = \frac{1}{4\sqrt{2}} \frac{(\sqrt{3/4})^{2}}{\frac{1}{2}\sqrt{\frac{1}{2}}} $

$= \frac{(\sqrt{3/4})^{2}}{2\sqrt{2} \, \pi}$