TO calculate the continious integral of : 1/(1 + cos(theta))

You need to use the substitution `cos theta = (1 - y^2)/(1
+ y^2)`

`y = tan (theta/2) =gt dy = (d
theta)/(2cos^2(theta/2)) =gt d theta = (2(1 - y^2)dy)/(1 +
y^2)` 

`int (d theta)/(1 + cos theta) = int (2(1 -
y^2)dy)/(1 + y^2)/((1 + y^2 +1 - y^2)/(1 + y^2))`

`` `int
(2(1 - y^2)dy)/(1 + y^2 +1 - y^2) = 2int ((1 -
y^2)dy)/2`

`2int ((1 - y^2)dy)/2 = int dy - int y^2 dy = y
- y^3/3 + c`

You need to come back to the original variable
such that:

`int (d theta)/(1 + cos theta) = tan (theta/2) -
(tan^3 (theta/2))/3 + c`

Hence, evaluating
the given integral yields:`int (d theta)/(1 + cos theta) = tan (theta/2) - (tan^3
(theta/2))/3 + c`