Question

不定积分问题

求定积分问题 ∫0到π [xsinx/(1+sinx)]dx

Answer 1

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