Question

数学问题 ∫x(sinx)^3dx 求不定积分求大神指点……

Answer 1

由∫f(x)g(x)dx=f(x)∫g(x)dx - ∫f '(x)g(x)dx有
P=∫x(sinx)^3dx=sin^3x·x^2/2-∫3xsin^2xcosxdx=sin^3x·x^2/2-3S

=x^2sin^3x/2-[3/2x^2sin^2xcosx-∫3x(2sinxcos^2x-sin^3x)]
=x^2sin^3x/2-3/2x^2sin^2xcosx+6T-3P
Q=∫x(cosx)^3dx=cos^3x·x^2/2+∫3xcos^2xsinxdx=cos^3x·x^2/2+3T

=x^2cos^3x/2+[3/2x^2cos^2xsinx+∫3x(2cosxsin^2x-cos^3x)]
=x^2cos^3x/2+3/2x^2cos^2xsinx+6S-3Q
则有
P=sin^3x·x^2/2-3S =x^2sin^3x/2-3/2x^2sin^2xcosx+6T-3P
Q=cos^3x·x^2/2+3T=x^2cos^3x/2+3/2x^2cos^2xsinx+6S-3Q
解方程即可解得P、Q
P即为所求

Answer 2

∫x(sinx)^3dx
=∫xsinx(1-(cosx)^2)dx
=∫xsinxdx-∫xsinx(cosx)^2)dx
=-∫xdcosx+1/3∫xd(cosx)^3
=-xcosx+∫cosxdx+1/3(x(cosx)^3-∫(cosx)^3dx
=-xcosx+sinx+1/3x(cosx)^3-1/3∫(1-(sinx)^2)dsinx
=-xcosx+sinx+1/3x(cosx)^3-1/3sinx+1/9(sinx)^3+C
=1/9(sinx)^3+1/3x(cosx)^3+2/3sinx-xcosx+C

Answer 3

∫xsin³x dx
= ∫x(1-cos²x)sinx dx
= ∫xsinx dx - ∫xcos²xsinx dx
= ∫xsinx dx - (1/2)∫x(1+cos2x)sinx dx
= ∫xsinx dx - (1/2)∫xsinx dx - (1/2)∫xsinxcos2x dx
= (1/2)∫xsinx dx - (1/4)∫x(sin3x-sinx) dx
= (1/2)∫xsinx dx - (1/4)∫xsin3x dx + (1/4)∫xsinx dx
= -(3/4)∫x dcosx + (1/4)(1/3)∫x dcos3x
= -(3/4)xcosx + (3/4)∫cosx dx + (1/12)xcos3x - (1/12)∫cos3x dx
= -(3/4)xcosx + (3/4)sinx + (1/12)xcos3x - (1/36)sin3x + C

Answer 4

∫x(sinx)^3dx
＝1/2 X (sinX)³＋1/4(COSX)^4