Given: Find #int(1/(1+cot z))dz#
Let #u = tan z; " "1/u = cot z; " "du = sec^2z dz; " " dz = (du)/(sec^2 z)#
Use the Pythagorean Identity: #tan^2 z + 1 = sec^2 z; " "u^2 + 1 = sec^2 z#
so #dz = (du)/(u^2+1)#
so #int(1/(1+cot z))dz = int(1/(1+1/u))((du)/(u^2+1)) = int(1/((u+1)/u))(1/(u^2+1))du = int(u/((u+1)(u^2+1)))du#
Use Partial Fractions to simplify the integrand:
#u/((u+1)(u^2+1)) = A/(u+1) + (Bu+C)/(u^2+1)#
Use common denominator to get an equivalent numerator equation:
#u/((u+1)(u^2+1)) = A/(u+1) * (u^2+1)/(u^2+1) + (Bu+C)/(u^2+1)*(u+1)/(u+1)#
#u = A(u^2+1) + (Bu + C)(u+1)#
Distribute: #" "u = Au^2 + A + Bu^2 + Bu + Cu + C#
Regroup: #" "(A+B)u^2 + (B+C)u + (A+C)#
#A + B = 0; " " A = -B#
#B + C = 1; " " C = 1-B#
#A + C = 0 => " use substitution :" -B + 1-B = 0#
#-2B + 1 = 0; -2B = -1; " "B = 1/2#
#A = -1/2; " " C = 1 - 1/2 = 1/2#
so, #u/((u+1)(u^2+1)) = -1/(2(u+1)) + (u+1)/(2(u^2+1))#
so #int(u/((u+1)(u^2+1)))du = int -(du)/(2(u+1))+int (u+1)/(2(u^2+1))du#
#= 1/2 int(u+1)/(u^2+1)du - 1/2 int(du)/(u+1)#
Expand first integrand: #" "1/2( int u/(u^2+1)du + int(du)/(u^2+1))- 1/2 int(du)/(u+1)#
#= 1/2 int u/(u^2+1)du + 1/2 int (du)/(u^2+1) - 1/2 int(du)/(u+1)#
Let #a = u^2 + 1; " "da = 2u du; " "du = (da)/(2u)#
For first integrand:
#1/2 int u/(u^2+1)du = 1/2 int cancel(u)/a (da)/(2cancel(u)) = 1/4 int (1/a)da = 1/4 ln (a) + C = 1/4 ln (u^2 + 1) + C = 1/4 ln (sec^2 z) + C#
For second integrand:
#1/2 int (du)/(u^2+1) = 1/2 tan^(-1) u + C #
#= 1/2 tan^(-1)(tan z) + C = z/2 + C#
For third integrand:
Let #p = u + 1; dp = du#
#-1/2 int(du)/(u+1) = -1/2 int 1/p dp = -1/2 ln (u+1) + C = -1/2 ln (tanz + 1) + C#
Combine all of the answers to all of the pieces:
#int(1/(1+cot z))dz = 1/4 ln (sec^2 z) + z/2 -1/2 ln (tanz + 1) + C#
#= 1/4 [ln (sec^2 z) + 2z -2ln (tanz + 1)] + C#
