$P=\int\limits_{0}^{\frac{\pi}{2}}sin^5xdx=\int\limits_{0}^{\frac{\pi}{2}}sin^4x(sinxdx)$$=\int\limits_{0}^{\frac{\pi}{2}}(1-cos^2x)^2d(-cosx)$
Đặt $u=-cosx\in [-1,0]$
$P=\int\limits_{-1}^{0}(1-u^2)^2du$
$=\int\limits_{-1}^{0}(1-2u^2+u^4)du$
$=u-\frac{2}{3}u^3+\frac{1}{5}u^5|_{-1}^0$
$=\frac{8}{15}$