$\int_{-\pi/2}^{\pi/2} dx \, \sin^{2n} x $

Since $$\int_{-\pi/2}^{\pi/2} e^{2ni\theta}\,d\theta = 0 $$ for any $n\in\mathbb{Z}\setminus\{0\}$, we have, by expanding $(e^{ix}-e^{-ix})^{2n}$ with the binomial theorem, $$I=\int_{-\pi/2}^{\pi/2}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n}\,dx = \frac{(-1)^n}{4^n}\binom{2n}{n}(-1)^n\frac{\pi}{2}=\frac{\pi}{2\cdot 4^n}\binom{2n}{n}.$$

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If someone is looking for an overkill, by using the Euler Beta function we have: $$ I = 2\int_{0}^{1} t^{2n}(1-t^2)^{-1/2}\,dt = \int_{0}^{1} u^{n-1/2}(1-u)^{-1/2}\,du = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}.$$