The integral \(\int_{0}^{\pi} (\sin(x)^{2n}) dx\) can be evaluated using integration by parts, leading to a recursive formula that relates \(\int (\sin(x)^{2n}) dx\) to \(\int (\sin(x)^{2n-1}) dx\). This method involves differentiating \(\sin(x)^{2n-1}\) and integrating a sine function. A closed-form solution exists and can be derived using the properties of the sine function and the Beta function, specifically \(B(n + 1/2, 1/2)\). The final result is \(\frac{\pi}{2^{2n}} \binom{2n}{n}\).
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