∫(0 π/2)(sinx)^8dx=(8-1)!/8!*(π/2)是怎么得到的?

有两种情况：
対于x∈[0,π/2],有
∫(0,π/2) sinⁿx dx = ∫(0,π/2) cosⁿx dx
= (n - 1)!/n!,n ≥ 3并且是奇数
= (n - 1)!/n!* π/2,n ≥ 2并且是偶数
证明,希望你能看明白吧,尤其是双阶乘的定义：
设f(n) = ∫(0,π/2) sinⁿx dx
= - ∫(0,π/2) sinⁿ⁻¹x d(cosx)
= - [sinⁿ⁻¹xcosx] |(0,π/2) + ∫(0,π/2) cosx d(sinⁿ⁻¹x)
= (n - 1)∫(0,π/2) sinⁿ⁻²xcos²x dx
= (n - 1)∫(0,π/2) sinⁿ⁻²x(1 - sin²x) dx
= (n - 1)[f(n - 2) - f(n)],循环
f(n) + (n - 1)f(n) = (n - 1)f(n - 2)
==> f(n) = (n - 1)/n * f(n - 2),n ≥ 2
= (n - 1)/n * (n - 3)/(n - 2) * f(n - 4)
= (n - 1)/n * (n - 3)/(n - 2) * (n - 5)/(n - 4) * f(n - 6)
= ...
当n是奇数：
则f(n) = [(n - 1)(n - 3)(n - 5) ...4 * 2]/[(n(n - 2)(n - 4) ...5 * 3 * 1] * f(1)
= [(n - 1)(n - 3)(n - 5) ...4 * 2]/[(n(n - 2)(n - 4) ...5 * 3 * 1] * 1
= (n - 1)!/n!
当n是偶数：
则f(n) = [(n - 1)(n - 3)(n - 5) ...3 * 1]/[n(n - 2)(n - 4) ...4] * f(2)
= (n - 1)(n - 3)(n - 5) ...3 * 1]/[n(n - 2)(n - 4) ...4 * 2] * π/2
= (n - 1)!/n!* π/2