高数不定积分
设f(x)原函数F(x)>0且F(0)=0当x≥0时f(x)F(x)=sin^2(2x)求f(x)
∫ ƒ(x) dx = F(x)
F'(x) = ƒ(x)
ƒ(x)F(x) = sin²2x = (1 - cos4x)/2两边积分
∫ ƒ(x)F(x) dx = ∫ [1/2 - (1/2)cos4x] dx
∫ F(x) d[F(x)] = ∫ [1/2 - (1/2)cos4x] dx
(1/2)[F(x)]² = x/2 - (1/8)sin4x C
[F(x)]² = x - (1/4)sin4x C
F(x) = √[x - (1/4)sin4x C]
F(0) = 0 ==> √[0 - 0 C] = 0 ==> C = 0
F(x) = √[x - (1/4)sin4x]
F'(x) = 2sin²2x/√(4x - sin4x) = ƒ(x)。
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