Question

求定积分上限0 下限π/2 （e^-x）乘以sin2xdx？

Answer 1

∫[π/2,0]e^-x * sin2x dx
= (-1/2)∫[π/2,0]e^-x dcos2x
= (1/2)∫[0,π/2]e^-x dcos2x
= (1/2)[e^-x * cos2x] - (1/2)∫[0,π/2]cos2x de^-x
= (1/2)[e^(-π/2) * (-1) - 1] + (1/4)(∫[0,π/2]e^-x dsin2x
= (1/2)[-e^(-π/2)-1] + (1/4)[e^-x * sin2x] + (1/4)∫[0,π/2]e^-x * sin2x dx
= (1/2)[-e^(-π/2)-1] - (1/4)∫[π/2,0]e^-x * sin2x dx
(1+1/4)∫[π/2,0]e^-x * sin2x dx = -(1/2)[e^(-π/2)+1]
∫[π/2,0]e^-x * sin2x dx = -(2/5)[e^(-π/2)+1],5,用两次分部积分就可以了,0,如图所示： ,0,