Fairly straightforward

Let $x = \dfrac{1}{\sqrt{2}} \sin t$ , then $dx = \dfrac{1}{\sqrt{2}} \cos t$

The integral becomes

$\displaystyle \int_{0}^{\frac{\pi}{4}} \sqrt{\frac{1}{2}} ( \frac{1}{2} \cos^2 t)^{(\frac{3}{2})} \cos t \hspace{4pt} dt$

$= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{4}} \cos^4 t \hspace{4pt} dt$

$= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{4}} ( \frac{1}{2} (1 + \cos( 2 t) ) )^2 \hspace{4pt} dt$

$= \frac{1}{16} \displaystyle \int_{0}^{\frac{\pi}{4}} (1 + 2 \cos( 2 t) + \cos^2 (2t) ) \hspace{4pt} dt$

$= \frac{1}{16} \displaystyle \int_{0}^{\frac{\pi}{4}} (1 + 2 \cos( 2 t) + \frac{1}{2} (1 + \cos( 4 t) ) )\hspace{4pt} dt$

$= \frac{1}{16} \displaystyle \int_{0}^{\frac{\pi}{4}} (\frac{3}{2} + 2 \cos( 2 t) + \frac{1}{2} \cos( 4 t) ) \hspace{4pt} dt$

$= \frac{1}{16} ( \frac{3}{2} t + \sin( 2 t) + \frac{1}{8} \sin( 4 t) ) \large \Big|_{0}^{\frac{\pi}{4}}$

$= \frac{1}{16} ( \frac{3}{2}( \frac{\pi}{4}) + 1+0 )$

$= \dfrac{3 \pi}{128} + \dfrac{1}{16}$