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这两个问题的积分,首先要做的就是降次.
(sin x)^2 = (1 - cos[2x]) / 2.
∴ ∫ (sin x)^2 dx = ∫ (1 - cos[2x]) / 2 dx = x/2 - 1/2 * ∫ cos[2x] dx = x/2 - 1/4 * sin[2x] + C.
cos[3x] = cos[2x + x] = cos[2x] * cos[x] - sin[2x] * sin[x]
= (2(cos[x])^2 - 1) * cos[x] - 2(sin[x])^2 * cos[x]
= 2 * (cos[x])^3 - cos[x] - 2 (1 - (cos[x])^2) * cos[x]
= 4 * (cos[x])^3 - 3 * cos[x]
∴ (cos[x])^3 = 3/4 * cos[x] + 1/4 * cos[3x]
∫ (cos[x])^3 dx = ∫ ( 3/4 * cos[x] + 1/4 * cos[3x] ) dx = 3/4 * sin[x] + 1/12 * sin[3x] + C.
有疑问可以追问~~
(sin x)^2 = (1 - cos[2x]) / 2.
∴ ∫ (sin x)^2 dx = ∫ (1 - cos[2x]) / 2 dx = x/2 - 1/2 * ∫ cos[2x] dx = x/2 - 1/4 * sin[2x] + C.
cos[3x] = cos[2x + x] = cos[2x] * cos[x] - sin[2x] * sin[x]
= (2(cos[x])^2 - 1) * cos[x] - 2(sin[x])^2 * cos[x]
= 2 * (cos[x])^3 - cos[x] - 2 (1 - (cos[x])^2) * cos[x]
= 4 * (cos[x])^3 - 3 * cos[x]
∴ (cos[x])^3 = 3/4 * cos[x] + 1/4 * cos[3x]
∫ (cos[x])^3 dx = ∫ ( 3/4 * cos[x] + 1/4 * cos[3x] ) dx = 3/4 * sin[x] + 1/12 * sin[3x] + C.
有疑问可以追问~~
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