Let $\sqrt{\cos t}=:u$，then $t=\arccos u^2$，${\rm d}t=-\dfrac{2u}{\sqrt{1-u^4}}{\rm d}u$. Therefore
\begin{align*} \int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t&=2\int_0^{1}\frac{u^6}{\sqrt{1-u^4}}{\rm d}u, \end{align*}
which seems to be the elliptic integral...
Edit
\begin{align*} \int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t&=\int_0^{\frac{\pi}{2}}\sin^0 t \cos^{\frac{5}{2}}t{\rm d}t\\ &=\frac{1}{2}B\left(\frac{1}{2},\frac{7}{4}\right)\\ &=\frac{1}{2}\cdot\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(\frac{9}{4}\right)}\\ &=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(\frac{9}{4}\right)}.\\ \end{align*}
Solution
Consider integrating by part. One can obtain \begin{align*} \int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t&=\int_0^{\frac{\pi}{2}}\cos^{\frac{3}{2}}t{\rm d}(\sin t)\\ &=\left[\cos^{\frac{3}{2}}t\sin t\right]_0^{\frac{\pi}{2}}+\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos^{\frac{1}{2}}t\sin^2 t{\rm d}t\\ &=0+\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos^{\frac{1}{2}}t(1-\cos^2 t){\rm d}t\\ &=\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos ^{\frac{1}{2}}t{\rm d}t-\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t. \end{align*} Thus $$\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t=\frac{3}{5}\int_0^{\frac{\pi}{2}}\cos ^{\frac{1}{2}}t{\rm d}t=\frac{3}{5}\int_0^{\frac{\pi}{2}}\sin ^{\frac{1}{2}}t{\rm d}t.$$
Let $u:=\sin^2 t$. Then ${\rm d} t=\dfrac{{\rm d}u}{2\sin t\cos t}.$ Hence \begin{align*} \int_0^{\frac{\pi}{2}}\sin ^{\frac{1}{2}}t{\rm d}t&=\int_0^1\frac{u^{\frac{1}{4}}}{2u^{\frac{1}{2}}(1-u)^{\frac{1}{2}}}{\rm d}u\\ &=\frac{1}{2}\int_0^1 u^{-1/4}(1-u)^{-1/2}{\rm d}u\\ &=\frac{1}{2}{\rm B}\left(\frac{3}{4},\frac{1}{2}\right)\\ &=\frac{\Gamma(3/4)\Gamma(1/2)}{2\Gamma(5/4)}\\ &=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma(3/4)}{\Gamma(5/4)}. \end{align*} It follows that $$\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t=\frac{3\sqrt{\pi}}{10}\cdot\frac{\Gamma(3/4)}{\Gamma(5/2)}\approx 0.71888\cdots$$