定积分的换元法应该怎样用?比如;x属于【1,4】,求(4-x^2)^(1/2)的dingjifen

问题描述:

定积分的换元法应该怎样用?
比如;x属于【1,4】,求(4-x^2)^(1/2)的dingjifen

∫[1 3lnx (lnx)^2]dx/x=∫[1 3lnx (lnx)^2]d(lnx)
lnx=t
原式=∫(1 3t t^2)dt=lnx 3(lnx)^2/2 (lnx)^3/3 C

积分限是有问题的,当x大于2时,是不成立的应该是x在(1,2)上为x/2(4-x^2)^1/2+2arcsinx/2
x的区间为(1,2)才行啊

∫√(a^2-x^2)dx
=a^2∫√[1-(x/a)^2]d(x/a)
x/a=cosu sinu=√(a^2-x^2)/a sin2u=2sinucosu=2x√(a^2-x^2)/a^2
=a^2∫√[1-(cosu)^2]dcosu
=a^2∫ -(sinu)^2du
=a^2∫[(cos2u-1)/2]du
=a^2(sin2u/2-u/2+C)
=a^2*[x√(a^2-x^2)/a^2-arccos(x/a)/2 +C0]
=x√(a^2-x^2)-(a^2/2)arccos(x/a)+C
a=2
∫[1,2] √(4-x^2)dx
= -1*√3+2*(π/3)