Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 58

Given to solve ,
int cos(theta) / (1+cos(theta)) d theta
just for easy solving let x=theta
so the equation is given as
int cos(x) / (1+cos(x)) d x -------(1)
let u= tan(x/2) ,=> then cos(x) is given as
=> cos(x) = (1-tan^2(x/2))/(1+tan^2(x/2)) = (1-u^2)/(1+u^2)
=>cos(x)= (1-u^2)/(1+u^2)
so dx = 2/(1+u^2) du
the expalnation is given below after this solution at NOTE.
so ,on substituting the value of u in the function (1) , we get
int cos(x) / (1+cos(x)) d x
=int ( (1-u^2)/(1+u^2)) / ( (1-u^2)/(1+u^2) +1) 2/(1+u^2) du
=int ( (1-u^2)/(1+u^2)) / ( (1-u^2+1+u^2)/(1+u^2) ) 2/(1+u^2) du
=int ( (1-u^2) / ( (1-u^2+1+u^2) ) 2/(1+u^2) du
=int ( (1-u^2) / ( (2) )) 2/(1+u^2) du
=int ( (1-u^2)/(1+u^2) du
=int ( (2-1-u^2)/(1+u^2) du
=int ((2)/(1+u^2)) -1 du
=int ((2)/(1+u^2)) du -int 1 du
=2int ((1)/(1+u^2)) du -u
as we know int ((1)/(1+u^2)) du = tan^(-1) u
so,
2int ((1)/(1+u^2)) du -u
=2 tan^(-1) u - u
but u= tan(x/2) ,so
= 2tan^(-1) (tan(x/2)) - tan(x/2) +c
= 2(x/2) - tan(x/2) +c
but x= theta ,so
= 2(theta/2) - tan(theta/2) +c
=theta - tan(theta/2) +c is the final answer

NOTE:

Explanation for cos(x) = (1-u^2)/(1+u^2)
before that , we know
cos(2x)= cos^2(x) -sin^2(x)
as cos^2(x) can be written as 1/(sec^2(x))
and we can show sin^2(x) = ((sin^2(x))/(cos^2(x) ))/(1/(cos^2(x)))
= tan^2(x)/sec^2x
so now ,
cos(2x)= cos^2(x) -sin^2(x)
= (1/sec^2(x)) - (tan^2(x)/sec^2(x))
=(1-tan^2(x))/(sec^2(x))
but sec^2(x) = 1+tan^2(x) ,as its an identity
so,
=(1-tan^2(x))/(sec^2(x))
=(1-tan^2(x))/(1+(tan^2(x)))

so ,
cos(2x) = (1-tan^2(x))/(1+(tan^2(x)))
so,
then
cos(x) = (1-tan^2(x/2))/(1+(tan^2(x/2)))
as before we told to assume that u= tan(x/2),
so,
cos(x) = (1-u^2)/(1+u^2)