Answer
See the explanation.
Work Step by Step
We need the reduction formula.
$\int_0^{\pi/2}\sin^{2n}x dx=\frac{2n-1}{2n}\int_0^{\pi/2}\sin^{2n-2} x dx$ (By the reduction formula)
$=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\int_0^{\pi/2}\sin^{2n-4} x dx$ (By the reduction formula)
$=\cdots$ (Continue the process)
$=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \ldots \frac{3}{4}\cdot \frac{1}{2}\int_0^{\pi/2}\sin^0 x dx$
$=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\int_0^{\pi/2} 1dx$
$=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\cdot [x]_0^{\pi/2}$
$=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\cdot (\frac{\pi}{2}-0)$
$=\frac{1\cdot 3\ldots (2n-3)(2n-1)}{2\cdot 4\ldots (2n-2)(2n)}\cdot \frac{\pi}{2}$