Question

求不定积分。谢谢了

Answer 1

2. ∫ xtan(x²) dx
=(1/2)∫ sin(x²)/cos(x²) d(x²)
=-(1/2)∫ 1/cos(x²) d(cosx²)
=-(1/2)ln|cos(x²)| + C

追答

1. ∫dx/(1+√(1-x^2))
x=sinu dx=cosudu √(1-x^2)=cosu
tan(u/2)=sinu/(1+cosu)=x/(1+√(1-x^2))
=∫cosudu/(1+cosu)
=∫[1-1/(1+cosu)]du
=u-∫du/(1+cosu)
=u-∫d(u/2)/(cos(u/2))^2
=u-tan(u/2)+C
=arcsinx - x/(1+√(1-x^2)) +C

追问

是(tanx)^2,不是tan(x^2)

追答

看错了，哈哈

∫x(tanx)^2dx=∫x[(secx)^2-1]dx=∫x(secx)^2dx-∫xdx
=∫xd(tanx)-(x^2)/2=-(x^2)/2+[xtanx-∫tanx dx]
=-(x^2)/2+[xtanx+∫(1/cosx)d(cosx)
=xtanx+ln|cosx|-(x^2)/2+C

Answer 2

let
x=sinu
dx = cosu du
∫ dx/[1+√(1-x^2)]
∫ [√(1-x^2)-1]/(-x^2) dx
=-1/x -∫ [√(1-x^2)]/x^2 dx
=-1/x -∫ (cotu)^2 du
=-1/x -∫ [(cscu)^2-1 ] du
=-1/x + cotu + u +C
=-1/x + √(1-x^2)/x + arcsinx + C
(2)
∫x(tanx)^2 dx
=∫x[(secx)^2-1] dx
=∫ xdtanx - (1/2)x^2
=xtanx -∫ tanx dx -(1/2)x^2
=xtanx + ln|cosx| -(1/2)x^2 + C

Answer 3

望能帮到你