✔ 最佳答案
Let In = ∫(x = 0 → π/2) sinn x dx
= -∫(x = 0 → π/2) sinn-1 x d(cos x)
= - [sinn-1 x cos x](x = 0 → π/2) + ∫(x = 0 → π/2) cos x d(sinn-1 x)
= (n - 1)∫(x = 0 → π/2) sinn-2 x cos2 x dx
= (n - 1)∫(x = 0 → π/2) (sinn-2 x - sinn x) dx
= (n - 1) (In-2 - In)
nIn = (n - 1)In-2
Hence
In = [(n - 1)/n] In-2
= [(n - 1)(n - 3)]/[n(n - 2)] In-4
.
.
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= [(n - 1)(n - 3) ... 4 x 2]/[n(n - 2) ... x 3 x 1] ∫(x = 0 → π/2) sin x dx if n is odd
= [(n - 1)(n - 3) ... 4 x 2]/[n(n - 2) ... x 3 x 1]
= (n - 1)!!/n!!
OR
= [(n - 1)(n - 3) ... 5 x 3]/[n(n - 2) ... x 6 x 4] ∫(x = 0 → π/2) sin2 x dx if n is even
= [(n - 1)(n - 3) ... 5 x 3]/[n(n - 2) ... x 6 x 4] x (1/2)∫(x = 0 → π/2) (1 - cos 2x) dx
= [(n - 1)(n - 3) ... 5 x 3]/[n(n - 2) ... x 6 x 4 x 2] x π/2
= [(n - 1)!!/n!!] (π/2)
For ∫(x = 0 → π/2) cosn x dx, sub u = π/2 - x, then dx = -du
∫(x = 0 → π/2) cosn x dx = -∫(u = π/2 → 0) sinn u du
= ∫(u = 0 → π/2) sinn u du
= ∫(x = 0 → π/2) sinn x dx since u is a dummy variable
Hence ∫(x = 0 → π/2) cosn x dx = ∫(x = 0 → π/2) sinn x dx for any n >= 2