计算∫0π（0下标,π上标）（1-cos ³x）dx

∫[0→π] (1-cos³x) dx
=∫[0→π] 1 dx - ∫[0→π] cos³x dx
=π - ∫[0→π] cos²x dsinx
=π - ∫[0→π] (1-sin²x) dsinx
=π - [sinx - (1/3)sin³x] |[0→π]
=π
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