高数 定积分 这是怎么得来的呀? 10
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因 ∫<0, π/2>sinxdx = [-cosx]<0, π/2> = 1,
∫<0, π/2>cosxdx = [sinx]<0, π/2> = 1.
故得 ∫<0, π/2>sinxdx = ∫<0, π/2>cosxdx。
I = ∫<0, π>(sinx)^3dx = ∫<0, π/2>(sinx)^3dx + ∫<π/2, π>(sinx)^3dx
后者令 x = π/2+u, 则
I2 = ∫<π/2, π>(sinx)^3dx = ∫<0, π/2>(cosu)^3du = ∫<0, π/2>(sinu)^3du
I = ∫<0, π/2>(sinx)^3dx + ∫<0, π/2>(sinu)^3du = 2∫<0, π/2>(sinx)^3dx, 即得。
∫<0, π/2>cosxdx = [sinx]<0, π/2> = 1.
故得 ∫<0, π/2>sinxdx = ∫<0, π/2>cosxdx。
I = ∫<0, π>(sinx)^3dx = ∫<0, π/2>(sinx)^3dx + ∫<π/2, π>(sinx)^3dx
后者令 x = π/2+u, 则
I2 = ∫<π/2, π>(sinx)^3dx = ∫<0, π/2>(cosu)^3du = ∫<0, π/2>(sinu)^3du
I = ∫<0, π/2>(sinx)^3dx + ∫<0, π/2>(sinu)^3du = 2∫<0, π/2>(sinx)^3dx, 即得。
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