Definite integration

2011-11-29 11:24 pm
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The method of substitution is suggested.

回答 (1)

2011-11-30 12:05 am
✔ 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
參考: 原創答案


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