Question

高等数学，求定积分

11和12

Answer 1

(11)
let
x=sinu
dx=cosu du
x=1/√2 , u=π/4
x=1, u=π/2
∫(1/√2->1) √(1-x^2)/x^2 dx
=∫(π/4->π/2) (cotu)^2 du
=∫(π/4->π/2) [ (cscu)^2 -1] du
=[-cotu -u]|(π/4->π/2)
=( 1+π/4) -π/2
=1 - π/4
(12)
let
x=tanu
dx=(secu)^2 du
x=1, u=π/4
x=√3, u=π/3
∫(1->√3) dx/[x^2.√(1+x^2)]
=∫(π/4->π/3) (secu)^2 du/[(tanu)^2.secu]
=∫(π/4->π/3) (secu)/(tanu)^2 du
=∫(π/4->π/3) (cosu)/(sinu)^2 du
=∫(π/4->π/3) dsinu/(sinu)^2
=-[1/sinu]|(π/4->π/3)
=√2 - 2/√3
=√2 - (2/3)√3

Answer 2

(11)

令x=sint，则dx= cost dt，积分范围由[1/√2,1]变为[π/4,π/2]
原积分
=∫cos²t/sin²t *dt
=∫(1/sin²t-1)*dt
=-cot(t)-t
= 0-π/2+1+π/4
=1-π/4
(12)
令x=tant，则dx=dt/(cos²t) 积分范围由[1,√3]变为[π/4,π/3]
原积分
=∫cost/sin²t *dt
= ∫1/sin²t *dsint
= -1/sint
= √2-2/√3
= √2-2√3/3

Answer 3

11题，令sint=x，dx=costdt
原式=cost/sint^2*costdt=cost^2/sint^2dt=[1/sint^2-1]dt