Question

求这十个题的不定积分

Answer 1

答案在纸上

Answer 2

1.

😁

追答

∫cscx dx
=∫1/sinx dx
=∫1/[2sin(x/2)cos(x/2)] dx,两倍角公式
=∫1/[sin(x/2)cos(x/2)] d(x/2)
=∫1/tan(x/2)*sec²(x/2) d(x/2)
=∫1/tan(x/2) d[tan(x/2)],注∫sec²(x/2)d(x/2)=tan(x/2)+C
=ln|tan(x/2)|+C,这是答案一

进一步化简:
=ln|sin(x/2)/cos(x/2)|+C
=ln|2sin(x/2)cos(x/2)/[2cos²(x/2)]|+C,凑出两倍角公式
=ln|sinx/(1+cosx)|+C
=ln|sinx(1-cosx)/sin²x|+C
=ln|(1-cosx)/sinx|+C
=ln|cscx-cotx|+C,这是答案二

8 ∫ xln(1+x)dx
=∫ln(1+x)dx²/2
=x²/2ln(1+x)-1/2∫x²/(1+x)dx
=x²/2 ln(1+x)-1/2∫(x²-1+1)/(1+x)dx
=x²/2 ln(1+x)-1/2∫[(x-1)+1/(1+x)]dx
=1/2x²ln(1+x)-1/2x²+1/2x-1/2ln(1+x)+c

∫e^2xcos3xdx
=1/3∫e^2xcos3xd3x
=1/3∫e^2xdsin3x
=1/3e^2xsin3x-1/3∫sin3xde^2x
=1/3e^2xsin3x-2/3∫sin3xe^2xdx
=1/3e^2xsin3x-2/9∫sin3xe^2xd3x
=1/3e^2xsin3x+2/9∫e^2xdcos3x
=1/3e^2xsin3x+2/9e^2xcos3x-2/9∫cos3xde^2x
=1/3e^2xsin3x+2/9e^2xcos3x-4/9∫e^2xcos3xdx
所以13/9∫e^2xcos3xdx=1/3e^2xsin3x+2/9e^2xcos3x
所以∫e^2xcos3xdx=3/13e^2xsin3x+2/13e^2xcos3x+C

10. ∫xarctan2xdx = (1/2)∫arctan2xdx^2
= (1/2)x^2arctan2x - (1/2)∫2x^2/(1+4x^2)dx
= (1/2)x^2arctan2x - (1/4)∫(1+4x^2-1)/(1+4x^2)dx
= (1/2)x^2arctan2x - x/4 +(1/8)∫1/(1+4x^2)d(2x)
= (1/2)x^2arctan2x - x/4 +(1/8)arctan2x + C

Answer 3