#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)))#.
