[微積] ∫sin^2(nx)dx 0...π
π
∫sin^2(nx)dx = π/2 for every n € N
0
為何我算出來卻是π/2n ...
令u=nx, du=n dx
第一步原式=
(1/n)∫sin^2(u) du
這樣錯了嗎Orz
感謝指教>"<
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