Example

Tag 089Y, permanent. Cite it as [Bourbaki, Tag 089Y].

One has

\sin^{2n} x = \frac{(-1)^n}{2^{2n}} \left( e^{i x} - e^{-i x} \right)^{2n} = \frac{(-1)^n}{2^{2n}} \left( e^{2n i x} - \binom{2n}{1} e^{(2n-2)i x} + \cdots + e^{-2n i x} \right)

whence

\int_0^1 \sin^{2n} t \, dt = \frac{(-1)^n}{2^{2n}} \left( \frac{1}{n} \sin 2n x - \binom{2n}{1} \frac{1}{n-1} \sin(2n-2)x + \cdots \right.

\left. + (-1)^{n-1} \binom{2n}{n-1} \sin 2x + (-1)^n \binom{2n}{n} x \right)

and in particular

\int_0^{\pi/2} \sin^{2n} t \, dt = \binom{2n}{n} \frac{1}{2^{2n}} \frac{\pi}{2} = \frac{1.3.5 \ldots (2n-1)}{2.4.6 \ldots 2n} \frac{\pi}{2}.

(29)

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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