Question

这个数学题怎么做啊？想了好久了，求助

Answer 1

设lnx＝t
所以x=e^t
t的范围是[0,1]
A＝∫sintde^t
＝e^tsint-∫e^tcostdt
＝e^tsint-∫costde^t
＝e^tsint-e^tcost-∫e^tsintdt
＝e^t(sint-cost)-A
所以A＝1/2e^t(sint-cost)
所以结果为1/2e(sin1-cos1)-1/2(0-1)
＝1/2e(sin1-cos1)+1/2

Answer 2

∫(1->e) sin(lnx) dx
=[xsin(lnx)]|(1->e) - ∫(1->e) cos(lnx) dx
=esin1 - [xcos(lnx)]|(1->e) -∫(1->e) sin(lnx) dx
2∫(1->e) sin(lnx) dx = esin1 -ecos1 +1
∫(1->e) sin(lnx) dx = (1/2) [e(sin1 -cos1) +1 ]