Determine the integral $\displaystyle \int^{\frac{\pi}{2}}_0 \cos^5 x dx$

$ \begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 \cos^4 x \cos x dx \qquad \text{Apply Trigonometric Identity } \cos^2x = 1 - \sin^2 x \\ \\ \int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 (1 - sin^2 x)^2 \cos x dx \\ \\ \int^{\frac{\pi}{2}}_0 \cos^5 x dx =& \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx \end{aligned} \end{equation} $

Let $u = \sin x$, then $du = \cos x dx$, when $x = 0, u = 0$ and when $\displaystyle x = \frac{\pi}{2}, u = 1$. Therefore,

$ \begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \int^1_0 (1 - 2u^2 + u^4) du \\ \\ \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& u - 2 \cdot \frac{u^{2 + 1}}{2 + 1} + \left. \frac{u^{4 + 1}}{4 + 1} \right|^1_0 \\ \\ \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& u - \frac{2u^3}{3} + \frac{u^5}{5} \left. \right|^1_0 \\ \\ \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& 1 - \frac{2(1)^3}{3} + \frac{(1)^5}{5} - 0 + \frac{2(0)^3}{3} - \frac{(0)^5}{5} \\ \\ \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \frac{15 - 10 + 3}{15} \\ \\ \int^{\frac{\pi}{2}}_0 (1 - 2 \sin^2 x + \sin^4 x) \cos x dx =& \frac{8}{15} \end{aligned} \end{equation} $