求定积分上限0 下限π/2 (e^-x)乘以sin2xdx
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求定积分上限0 下限π/2 (e^-x)乘以sin2xdx
∫[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]
= (-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]
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