Re: [積分] 定積分2
※ 引述《henry9621205 (清心小子)》之銘言:
: π (sin9x)^2
: ∫ ------------ dx
: 0 (sinx)^2
: 謝謝!!
---
π {sin[(2n+1)x]}^2
∫ ──────── dx , n屬於N
0 [sin(x)]^2
π n 2
= ∫ [ 1 + 2 Σ cos(2kx) ] dx
0 k=1
π n n n
= ∫ 1 + 4 Σ cos(2kx) + 4[ Σ cos(2kx)][ Σ cos(2kx) ] dx
0 k=1 k=1 k=1
nπ
= π + 0 + 4*──
2
= (2n+1)π
ps:
π
∫ cos(2mx)*cos(2nx) dx = ┌ π/2 if m=n , m、n 屬於N
0 └ 0 if m≠n
--
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