標題:
integration qq
發問:
? π/2 e^sinx sin2xdx 0 let u = sinx then...?
∫ (0 to π/2) e^sinx sin2xdx let u = sinx x=0 ---> u=sin0 =0 x=π/2-->u=sin(π/2)=1 du=cos x dx ∫ (0 to π/2) e^sinx sin2xdx = ∫ (0 to π/2) e^sinx* (2sin x cosx )dx = 2 ∫ (0 to π/2) sin x *e^sinx d(sin x ) =2 ∫ ( 0 to 1 )u *e^udu =2 ∫ (0 to 1 )u d(e^u) =2 [ u*e^u](0,1) - 2 ∫ (0 to 1)e^u du =2e^u(u- 1)------ in(0,1) =2 =2e- 2007-12-09 10:22:16 補充： sorry, the last line is not needed
其他解答:
integration qq
發問:
? π/2 e^sinx sin2xdx 0 let u = sinx then...?
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最佳解答:∫ (0 to π/2) e^sinx sin2xdx let u = sinx x=0 ---> u=sin0 =0 x=π/2-->u=sin(π/2)=1 du=cos x dx ∫ (0 to π/2) e^sinx sin2xdx = ∫ (0 to π/2) e^sinx* (2sin x cosx )dx = 2 ∫ (0 to π/2) sin x *e^sinx d(sin x ) =2 ∫ ( 0 to 1 )u *e^udu =2 ∫ (0 to 1 )u d(e^u) =2 [ u*e^u](0,1) - 2 ∫ (0 to 1)e^u du =2e^u(u- 1)------ in(0,1) =2 =2e- 2007-12-09 10:22:16 補充： sorry, the last line is not needed
其他解答:
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