\displaystyle I_n = \int \limits_0^{\pi / 2} \sin^n x \, dx = \left[
\begin{array}{ll}
u = \sin^{n - 1} x & du = (n - 1) \sin^{n - 2} x \cos x \, dx \\
dv = \sin x \, dx & v = -\cos x
\end{array} \right] \\ = \ldots = (n - 1) (I_{n - 2} - I_n) \ \Rightarrow \ I_n = \frac{n - 1}{n} I_{n - 2}, I_0 = \frac{\pi}{2}