Question

关于三角函数的不定积分.求大佬.

Answer 1

解：1-cosx=2cos^2(x/2). sinx=2sin(x/2)cos(x/2);
1-cosx+sinx=2cos(x/2)[sin(x/2)+cos(x/2)]
原式=∫[2sin(x/2)cos(x/2)]^3dx/[2sin(x/2)+cos(x/2)]
=∫4sin^3(x/2)[cos^2(x/2)-1/2+1/2]dx/[sin(x/2)+cos(x/2)]
=2∫sin^3(x/2)dx/[sin(x/2)+cos(x/2)]+2∫sin^3(x/2)[2cos^2(x/2)-1]dx/[sin(x/2)+cos(x/2)]
=∫sin(x/2)(1+1-2cos^2(x/2)]dx/[sin(x/2)+cos(x/2)]+2∫sin^3(x/2)[sin(x/2)-cos(x/2)]dx
=2∫sin(x/2)dx/[sin(x/2)+cos(x/2)]+[sin(x/2)+cos(x/2)]dx+2∫sin^4(x/2)dx+4sin^3(x/2)dsin(x/2)
=sin^4(x/2)-2√2∫dcos(x/2-π/4)/cos(x/2-π/4)]+2∫[sin(x/2)+cos(x/2)]d(x/2)+2∫sin^2(x/2)[1-cos^2(x/2)]dx+C1
=sin^4(x/2)-2√2ln|cos(x/2-π/4)|-2cos(x/2)+2sin(x/2)+2∫sin^2(x/2)dx-(1/2)sin^2xdx+C2;
=sin^4(x/2)-2√2ln|cos(x/2-π/4)|-2cos(x/2)+2sin(x/2)-sinx+x+(1/4)sin(2x)-(x/4)+C
=sin^4(x/2)-2√2ln|cos(x/2-π/4)|-2cos(x/2)+2sin(x/2)-sinx+(1/4)sin(2x)+(3x/4)+C