求定积分的问题
1个回答
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let
f(x)=x^3/(cosx)^2
f(-x)=-f(x)
=>∫(-π/4->π/4) x^3/(cosx)^2 dx =0
//
∫(-π/4->π/4) (1+x^3)/(cosx)^2 dx
=∫(-π/4->π/4) dx/(cosx)^2 +∫(-π/4->π/4) x^3/(cosx)^2 dx
=∫(-π/4->π/4) dx/(cosx)^2
=∫(-π/4->π/4) (secx)^2 dx
=[tanx]|(-π/4->π/4)
=2
f(x)=x^3/(cosx)^2
f(-x)=-f(x)
=>∫(-π/4->π/4) x^3/(cosx)^2 dx =0
//
∫(-π/4->π/4) (1+x^3)/(cosx)^2 dx
=∫(-π/4->π/4) dx/(cosx)^2 +∫(-π/4->π/4) x^3/(cosx)^2 dx
=∫(-π/4->π/4) dx/(cosx)^2
=∫(-π/4->π/4) (secx)^2 dx
=[tanx]|(-π/4->π/4)
=2
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