Question

定积分计算题

如图，lnx/x^2+a^2那题

Answer 1

令x=atant，则dx=asec^2tdt
原式=∫(0,π/2) ln(atant)/(a^2*sec^2t)*asec^2tdt
=(1/a)*∫(0,π/2) ln(atant)dt
=(1/a)*∫(0,π/2) [lna+ln(tant)]dt
=(1/a)*[tlna|(0,π/2)+∫(0,π/2) ln(tant)dt]
=(πlna)/(2a)+(1/a)*∫(0,π/2) ln(tant)dt
=(πlna)/(2a)+(1/a)*[∫(0,π/4) ln(tant)dt+∫(π/4,π/2) ln(tant)dt]
=(πlna)/(2a)+(1/a)*[∫(0,π/4) ln(tant)dt+∫(π/4,π/2) ln(cot(π/2-t))dt]
对最后一项，令u=π/2-t，则dt=-du
原式=(πlna)/(2a)+(1/a)*[∫(0,π/4) ln(tant)dt-∫(π/4,0) ln(cotu)du]
=(πlna)/(2a)+(1/a)*[∫(0,π/4) ln(tant)dt-∫(π/4,0) ln(1/tanu)du]
=(πlna)/(2a)+(1/a)*[∫(0,π/4) ln(tant)dt-∫(0,π/4) ln(tanu)du]
=(πlna)/(2a)

Answer 2

(1)
∫(1->e^2) dx/[x√lnx]
=∫(1->e^2) dlnx/√lnx
=2[√lnx]|(1->e^2)
=2√2
(5)
lim(x->0) ∫(0->x^2) tsint dt/∫(0->sinx) t^5 dt (0/0 分子分母分别求导)
=lim(x->0) 2x.[x^2.sin(x^2)]/ [ cosx.(sinx)^5 ]
=lim(x->0) 2x^5/ [ cosx. (x^5) ]
=lim(x->0) 2/cosx
=2

Answer 3

∫(1->e^2) dx/[x√lnx]

=∫(1->e^2) dlnx/√lnx

=2[√lnx]|(1->e^2)

=2√2