Question

大一 高数 定积分

Answer 1

原式=∫(0,π/2)√cosx(1-cos²x)dx
=∫(0,π/2)√cosxsin²x dx
=∫(0,π/2)√cosx· sinxdx
=-∫(0,π/2)√cosxdcosx
=-1/(1/2+1) (cosx)^(1/2+1)|(0,π/2)
=-2/3 ·cosx^(3/2)|(0,π/2)
=2/3

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Answer 2

在 [0,π/2] 范围内，cosx ≥0, sinx≥0
所以，
=∫√[cosx * (1 - cos²x)] * dx
=∫√[cosx * sin²x] * dx
=∫√(cosx) * sinx * dx
=-∫√(cosx) * (-sinx * dx)
=-∫√(cosx) * d(cosx)
=-2/3 * √(cosx)³|x=0→π/2
=-2/3 * [√(cosπ/2)³ - √(cos0)³]
=-2/3 * [√0 - √1]
=2/3