Question

不定积分问题

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• 系统奖励15（财富值+成长值）+难题奖励20（财富值+成长值）

Answer 1

(2)、∫x²cosxdx=∫x²dsinx=x²sinx-∫sinxdx²=x²sinx-2∫xsinxdx=x²sinx+2∫xdcosx
=x²sinx+2xcosx-2∫cosxdx=x²sinx+2xcosx-2sinx+C
(6)、∫e^xcosxdx=∫e^xdsinx=e^xsinx-∫e^xsinxdx
=e^xsinx-∫e^(x+π/2)sin(x+π/2)d(x+π/2)=e^xsinx+e^(π/2)∫e^xcosxdx
所以：∫e^xcosxdx=e^xsinx/[1-e^(π/2)]+C
(17)、∫e^√(3x+9)dx
=(1/3)∫e^√(3x+9)d(3x+9) 设u=√(3x+9)
=(1/3)∫e^udu²
=(2/3)∫ude^u
=(2/3)ue^u-(2/3)∫e^udu
=(2/3)ue^u-(2/3)e^u+C
=(2/3)(u-1)e^u+C
=(2/3)(√(3x+9)-1)e^√(3x+9)+C

Answer 2

(2) ∫x^2cosxdx = ∫x^2dsinx = x^2sinx - ∫2xsinxdx
= x^2sinx + 2∫xdcosx = x^2sinx + 2xcosx - 2∫cosxdx
= x^2sinx + 2xcosx - 2sinx + C
(6) I = ∫e^x cosxdx = ∫e^xdsinx = e^xsinx - ∫e^x sinxdx
= e^xsinx + ∫e^xdcosx = e^xsinx + e^xcosx - ∫e^xcosxdx
= e^xsinx + e^xcosx - I
I = (1/2)e^x(sinx+cosx) + C
(17) 令 u = √(3x+9), 则 x = (u^2-9)/3, dx = (2/3)udu
I = ∫e^u (2/3)udu = (2/3)∫ude^u = (2/3)[ue^u - ∫e^udu]
= (2/3)(u-1)e^u + C = (2/3) [√(3x+9)-1] e^[√(3x+9)] + C