高数大一
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高数大一
答:
1)
∫ [2/(x-3)]dx
=∫ [2/(x-3)]d(x-3)
=2ln|x-3|+C
2)
(0→π/2) ∫ (cosx)^2dx
=(0→π/2) ∫ (cos2x+1)/2 dx
=(0→π/2) (1/4)*∫(cos2x+1)d(2x)
=(0→π/2) (sin2x+2x)/4
=(0+π)/4-0
=π/4
3)
∫ (tanx)^2 dx
=∫ [(sinx)^2/(cosx)^2] dx
=∫ {[1-(cosx)^2]/(cosx)^2} dx
=∫ [1/(cosx)^2-1]dx
=∫ [(secx)^2-1]dx
=tanx-x+C
此题应用到下面的公式:
1)
∫ [2/(x-3)]dx
=∫ [2/(x-3)]d(x-3)
=2ln|x-3|+C
2)
(0→π/2) ∫ (cosx)^2dx
=(0→π/2) ∫ (cos2x+1)/2 dx
=(0→π/2) (1/4)*∫(cos2x+1)d(2x)
=(0→π/2) (sin2x+2x)/4
=(0+π)/4-0
=π/4
3)
∫ (tanx)^2 dx
=∫ [(sinx)^2/(cosx)^2] dx
=∫ {[1-(cosx)^2]/(cosx)^2} dx
=∫ [1/(cosx)^2-1]dx
=∫ [(secx)^2-1]dx
=tanx-x+C
此题应用到下面的公式: