int 1/(1+bcot(x)) "d"x,
int tan(x)/(tan(x)+b) "d"x,
int (tan(x)+b-b)/(tan(x)+b) "d"x,
int 1 - b/(tan(x)+b) "d"x
x - b int 1/(tan(x)+b) "d"x.
Problem is reduced to the integral int 1/(tan(x)+b) "d"x.
Let u=tan(x). Then ("d"u)/("d"x) = sec^2(x). The trigonometric identity sec^2(x)=1+tan^2(x) gives "d"x = 1/(1+u^2) "d"u.
Integral is transformed to,
int 1/((1+u^2)(u+b)) "d"u.
Then perform a partial fraction decomposition.
Write
1/((1+u^2)(u+b))=(Au+C)/(1+u^2)+D/(u+b),
1/((1+u^2)(u+b))=(Au^2+Cu+Abu+Cb+D+Du^2)/((1+u^2)(u+b)),
1/((1+u^2)(u+b))=((A+D)u^2+(C+Ab)u+Cb+D)/((1+u^2)(u+b)).
Then by equating coefficients A=-D, A=-1/bC which gives D=1/b C. Then as Cb+D=1, C(b+1/b)=1, which gives C=b/(1+b^2).
This then readily gives D=1/(1+b^2), A=-1/(1+b^2).
The integral becomes,
1/(1+b^2) int (b-u)/(1+u^2) + 1/(u+b) "d"u
1/(1+b^2) ( b int 1/(1+u^2) "d"u - 1/2 int 2u/(1+u^2) "d"u + int 1/(u+b) "d"u).
Then by standard integration results the integral is,
1/(1+b^2) (b arctan(u) - 1/2lnabs(1+u^2) + lnabs(u+b)).
Substituting this back in and replacing u with tan(x) we have
int 1/(1+bcot(x)) "d"x = x - b/(1+b^2) (bx - 1/2 lnabs(1+tan^2(x)) + lnabs(tan(x)+b)),
int 1/(1+bcot(x)) "d"x = x (1-b^2/(1+b^2)) - b/(1+b^2) - 1/2 lnabs((sec(x))^2) + lnabs(tan(x)+b)),
int 1/(1+bcot(x)) "d"x = b/(1+b^2) (x/b + lnabs(sec(x) - ln(tan(x)+b)),
int 1/(1+bcot(x)) "d"x = 1/(1+b^2) (x - blnabs(sin(x)+bcos(x))).
