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Change integration variables:

$$ I(n) = \int_0^\pi \sin^{2n} x \, dx = - \int_{x=0}^{x=\pi} \sin^{2n-1} x \, d(\cos x)$$

Integrate by parts:

$$ I(n) = - \sin^{2n-1} x \cos x |_{x=0}^{x=\pi} + \int_{x=0}^{x=\pi} \cos x \, d(\sin^{2n-1} x) = (2n-1) \int_0^\pi \sin^{2n-2} x \, \cos^2 x \, dx $$

Use a trigonometric identity:

$$ I(n) = (2n-1) \int_0^\pi \sin^{2n-2} x \, (1 - \sin^2 x) \, dx = (2n-1) I(n-1) - (2n-1) I(n)$$

giving us

$$ I(n) = \frac{2n-1}{2n} I(n-1) = \frac{(2n)(2n-1)}{4n^2} I(n-1) $$

This gives us

$$ I(n) = \frac{(2n)!}{2^{2n} n! n!} I(0) $$

The final step is to find I(0). That is easy; it is π. Thus,

$$ I(n) = \frac{(2n)!}{2^{2n} n! n!} \pi $$