∫(0~π/2)(sinx)^3dx=?
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∫(0~π/2)(sinx)^3dx=?
∫(0~π/2)(sinx)^3dx=?
∫(0~π/2)(sinx)^3dx=?
∫(sinx)^3dx
=∫(sin²x)sinxdx
=-∫(1-cos²x)dcosx
=-cosx+(1/3)cos³x+C
=>
∫(0~π/2)(sinx)^3dx
=-cosx+(1/3)cos³x,(0~π/2)
=-(1+1/3)
=-4/3
Sin[x]^3 dx= -(1-Cos[x]^2)dCos[x]
所以
∫Sin[x]^3 dx = ∫(Cos[x]^2-1)dCos[x] = 1/3Cos[x]^3 - Cos[x] + C
用公式,答案:2/3
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