∫cosec(x)dx=∫(cot(x)+cosec(x))cosec(x)dx/(cot(x)+cosec(x)),
∫cosec(x)dx=-∫(-cot(x)-cosec(x))cosec(x)dx/(cot(x)+cosec(x)),
∫cosec(x)dx=-∫(-cot(x)cosec(x)-cosec2(x))dx(cosec(x)+cot(x)).
∫cosec(x)dx=-ln|cosec(x)+cot(x)|+C, where C is integration constant.
cosec(x)+cot(x)=1/sin(x)+cos(x)/sin(x)=(1+cos(x))/sin(x).
-ln|cosec(x)+cot(x)|=ln|1/(cosec(x)+cot(x))|=ln|sin(x)/(1+cos(x))|=ln|sin(x)|-ln|1+cos(x)|.
d3y/dx3+dy/dx=cosec(x)=
(d/dx)(d2y/dx2+y)=(d/dx)(∫cosec(x)dx), then integrating:
d2y/dx2+y=f(x), where f(x)=ln|sin(x)|-ln|1+cos(x)|+C.
d2y/dx2+y=0 has the solution yc (characteristic): Asin(x)+Bcos(x), where A and B are constants.
Now we need the particular solution yp.
y=yc+yp. Here we start using variation of parameters.
We already have yc=Asin(x)+Bcos(x), now we use y1=sin(x) and y2=cos(x), where:
yp=u1sin(x)+u2cos(x). u1 and u2 are functions of x and we find them using Wronskian determinants w, w1, w2, and Cramer's Rule for solving simultaneous equations.
w=
| sin(x) cos(x) |
| cos(x) -sin(x) | = -sin2(x)-cos2(x)=-1.
We have here y1 and y2 in the first row and their respective derivatives in the second row.
Now Cramer's Rule introducing f(x). (See above for definition of f(x).)
w1=
| 0 cos(x) |
| f(x) -sin(x) | = -f(x)cos(x).
w2=
| sin(x) 0 |
| cos(x) f(x) | = f(x)sin(x).
u1'=w1/w=f(x)cos(x); u2'=w2/w=-f(x)sin(x). Integrate to find u1 and u2:
u1=∫(ln|sin(x)|-ln|1+cos(x)|+C)cos(x)dx=∫(ln|sin(x)|cos(x)dx-∫ln|1+cos(x)|cos(x)dx+C∫cos(x)dx;
u2=∫(-ln|sin(x)|+ln|1+cos(x)|-C)sin(x)dx=∫(-ln|sin(x)|sin(x)dx+∫ln|1+cos(x)|sin(x)dx-C∫sin(x)dx.
Consider the integrals separately: u1=I+J+K, u2=L+M+N.
I=∫(ln|sin(x)|cos(x)dx. Let u=ln|sin(x)|, then eu=sin(x), eudu=cos(x)dx.
I=∫ueudu; let p=u, then dp=du; let dq=eudu, then q=eu (integration by parts).
I=pq-∫qdp, that is:
I=ueu-∫eudu=ueu-eu=sin(x)ln|sin(x)|-sin(x).
K=C∫cos(x)dx=Csin(x); N=-C∫sin(x)dx=Ccos(x).
M=∫ln|1+cos(x)|sin(x)dx. Let u=ln|1+cos(x)|, then eu=1+cos(x), eudu=-sin(x)dx.
So M=-ueu+eu=-(1+cos(x))ln|1+cos(x)|+1+cos(x).
J=-∫ln|1+cos(x)|cos(x)dx and L=∫(ln|sin(x)|(-sin(x))dx remain to be integrated.
For J, let u=ln|1+cos(x)|, then du=-sin(x)dx/(1+cos(x)); let dv=cos(x)dx, then v=sin(x).
J=-sin(x)ln|1+cos(x)|-∫sin2(x)dx/(1+cos(x))=
-sin(x)ln|1+cos(x)|-∫(1-cos2(x))dx/(1+cos(x))=
-sin(x)ln|1+cos(x)|-∫(1-cos(x)dx=
-sin(x)ln|1+cos(x)|-x+sin(x).
For L, let u=ln|sin(x)|, then du=cot(x)dx; let dv=-sin(x)dx, then v=cos(x).
L=uv-∫vdu, that is:
L=cos(x)ln|sin(x)|-∫cos2(x)dx/sin(x)=cos(x)ln|sin(x)|-∫(1-sin2(x))dx/sin(x),
L=cos(x)ln|sin(x)|-∫(cosec(x)-sin(x))dx,
L=cos(x)ln|sin(x)|+ln|cosec(x)+cot(x)|-cos(x)=cos(x)ln|sin(x)|+ln|1+cos(x)|-ln|sin(x)|-cos(x).
u1=I+J+K, u2=L+M+N.
u1=sin(x)ln|sin(x)|-sin(x)-sin(x)ln|1+cos(x)|-x+sin(x)+Csin(x),
u1=sin(x)ln|sin(x)|-sin(x)ln|1+cos(x)|-x+Csin(x).
u2=cos(x)ln|sin(x)|+ln|1+cos(x)|-ln|sin(x)|-cos(x)-(1+cos(x))ln|1+cos(x)|+1+cos(x)+Ccos(x),
u2=cos(x)ln|sin(x)|-ln|sin(x)|-cos(x)ln|1+cos(x)|+1+Ccos(x).
yp=u1sin(x)+u2cos(x)=
sin2(x)ln|sin(x)|-sin2(x)ln|1+cos(x)|-xsin(x)+Csin2(x)+
cos2(x)ln|sin(x)|-cos2(x)ln|1+cos(x)|-cos(x)ln|sin(x)|+cos(x)+Ccos2(x)=
ln|sin(x)|-ln|1+cos(x)|-xsin(x)-cos(x)ln|sin(x)|+cos(x)+C.
The full solution is:
y=Asin(x)+Bcos(x)+ln|sin(x)|-ln|1+cos(x)|-xsin(x)-cos(x)ln|sin(x)|+C. (A cos(x) term has been absorbed into the constant B.)