...∫㏑(1+x⊃2;)∫sin⊃2;x/cos⊃3;xdx ∫arctan√xdx ∫...

发布网友发布时间：2024-10-02 21:32

共2个回答

热心网友时间：2024-10-19 04:30

(1)令u=√x,x=u^2,dx=2udu
∫√xsin√x dx
=2∫u^2*sinu du
=-2∫u^2 d(cosu)
=-2(u^2*cosu-∫cosu du^2)
=-2(u^2*cosu-2∫ucosu du)
=-2(u^2)cosu+4∫u d(sinu)
=-2(u^2)cosu+4(usinu-∫sinu du)
=-2(u^2)cosu+4usinu+4cosu+C
=-2xcos√x+4√xsin√x+4cos√x+C=2(2-x)cos√x+4√xsin√x+C
(2)
∫sin^2x/cos^3x dx
=∫secxtan^2x dx
=∫tanx d(secx)
=secxtanx-∫secx d(tanx)
=secxtanx-∫sec^3x dx,∫sec^3x dx不详细求了
=secxtanx-(1/2*secxtanx+1/2*ln|secx+tanx|)+C
=(1/2)secxtanx-(1/2)ln|secx+tanx|+C
(3)u=√x,x=u^2,dx=2udu
∫arctan√x dx
=2∫uarctanu dx
=2∫arctanu d(u^2/2)
=2[u^2/2*arctanu-1/2*∫u^2 d(arctanu)]
=(u^2)arctanu-∫u^2/(1+u^2) du
=(u^2)arctanu-∫(1+u^2-1)/(1+u^2) du
=(u^2)arctanu-∫[1-1/(1+u^2+1] du
=(u^2)arctanu-(u-arctanu)+C
=xarctan√x-√x+arctan√x+C=(x+1)arcran√x-√x+C
(4)u=√(1+cosx),u^2=1+cosx,2udu=-sinxdx
∫√(1+cosx)/sinx dx
=2∫ 1/(u^2-2) du
=2*1/(2√2)*ln|(u-√2)/(u+√2)|+C
=(1/√2)ln|[√(1+cosx)-√2]/[√(1+cosx)+√2]|+C

热心网友时间：2024-10-19 04:25

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