Let $I = \int_{0}^{\pi/2}\sqrt{\sin x}\cos^{5}x~dx = \int_{0}^{\pi/2} (\sin x)^{1/2} \cos^4 x \cos x ~dx$

Rewrite $\cos^4 x$ as $(\cos^2 x)^2 = (1 - \sin^2 x)^2$

So, $I = \int_{0}^{\pi/2} (\sin x)^{1/2} (1 - \sin^2 x)^2 \cos x ~dx$

Let $t = \sin x$, then $dt = \cos x ~dx$. When $x = 0$, $t = 0$, and when $x = \pi/2$, $t = 1$.

Thus, $I = \int_{0}^{1} t^{1/2} (1 - t^2)^2 dt = \int_{0}^{1} t^{1/2} (1 - 2t^2 + t^4) dt$

$I = \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) dt$

$I = \left[ \frac{t^{3/2}}{3/2} - 2\frac{t^{7/2}}{7/2} + \frac{t^{11/2}}{11/2} \right]_{0}^{1}$

$I = \left[ \frac{2}{3}t^{3/2} - \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2} \right]_{0}^{1}$

$I = \frac{2}{3} - \frac{4}{7} + \frac{2}{11} = \frac{2(77) - 4(33) + 2(21)}{231} = \frac{154 - 132 + 42}{231} = \frac{64}{231}$