$\int\limits_{0}^{\frac{\pi }{2}}\cos^{5} x dx= \int\limits_{0}^{\frac{\pi }{2}}\cos^{4} xd(\sin x) $
$=\int\limits_{0}^{\frac{\pi }{2}}(1-\sin^2x)^2 d(\sin x) $
$=\int\limits_{0}^{\frac{\pi }{2}}(1-2\sin^2 x +\sin^4x) d(\sin x) $
$=\left[ {\sin x-\frac{2}{3}\sin^3 x +\ \frac{1}{5}\sin^5 x } \right] _{0}^{\frac{\pi }{2}} $
$=\frac{8}{15} $.