Use the parametric formula:
cosx = (1-tan^2(x/2))/(1+tan^2(x/2))
and substitute:
t= tan(x/2)
x = 2 arctan t
dx = (2dt)/(1+t^2)
we have:
int dx/(cosx+cosa) = 2 int dt/(1+t^2) 1/((1-t^2)/(1+t^2) +cosa)
int dx/(cosx+cosa) = 2 int dt/((1-t^2) +cosa(1+t^2)
int dx/(cosx+cosa) = 2 int dt/( (1+cosa) -t^2(1-cosa) )
Factor the denominator:
int dx/(cosx+cosa) = 2 int dt/(( sqrt(1+cosa) - tsqrt(1-cosa)) ( sqrt(1+cosa) + tsqrt(1-cosa) )
and use partial fractions decomposition:
1/(( sqrt(1+cosa) - tsqrt(1-cosa)) ( sqrt(1+cosa) + tsqrt(1-cosa) ) )= A/( sqrt(1+cosa) - tsqrt(1-cosa)) + B/( sqrt(1+cosa) + tsqrt(1-cosa))
A( sqrt(1+cosa) + tsqrt(1-cosa)) + B(sqrt(1+cosa) - tsqrt(1-cosa)) = 1
{ ( (A+B) sqrt(1+cosa) = 1) , ((A-B) sqrt(1-cosa) = 0):}
A=B= 1/(2sqrt(1+cosa))
int dx/(cosx+cosa) = 1/sqrt(1+cosa) int dt/( sqrt(1+cosa) - tsqrt(1-cosa)) +1/sqrt(1+cosa) int dt/( sqrt(1+cosa) + tsqrt(1-cosa))
Simplify the expression:
int dx/(cosx+cosa) = int dt/((1+cosa) - tsqrt((1-cosa)(1+cosa)) )+int dt/( (1+cosa) + tsqrt((1-cosa)(1+cosa))
int dx/(cosx+cosa) = int dt/((1+cosa) - tsqrt(1-cos^2a) )+int dt/( (1+cosa) + tsqrt(1-cos^2a))
Now:
sqrt(1-cos^2a) = abs sin a
however because of the symmetry of the expression we can ignore the absolute value, as changing sin a with -sin a leaves everything unchanged:
int dx/(cosx+cosa) = int dt/((1+cosa) - tsina )+int dt/( (1+cosa) + tsina)
int dx/(cosx+cosa) = 1/sina (int dt/((1+cosa)/sina - t )+int dt/( (1+cosa)/sina + t))
int dx/(cosx+cosa) = 1/sina (-ln((1+cosa)/sina - t )+ln( (1+cosa)/sina + t)) +C
Using now the properties of logarithms:
int dx/(cosx+cosa) = 1/sina ln (( (1+cosa)/sina + t) / ((1+cosa)/sina - t ))+C
int dx/(cosx+cosa) = 1/sina ln (( 1+cosa + tsina) / (1+cosa - t sina))+C
Undoing the substitution:
int dx/(cosx+cosa) = 1/sina ln (( 1+cosa + tan(x/2)sina) / (1+cosa - tan(x/2) sina))+C
Multiply by cos(x/2) numerator and denominator of the argument:
int dx/(cosx+cosa) = 1/sina ln (( cos(x/2)+cosacos(x/2) + sin(x/2)sina) / (cos(x/2)+cos(x/2)cosa - sin(x/2) sina))+C
int dx/(cosx+cosa) = 1/sina ln (( cos(x/2)+cos(x/2-a) ) / (cos(x/2)+cos(x/2+a)))+C
Use now the identity:
cos alpha + cos beta = 2cos((alpha+beta)/2)cos((alpha-beta)/2)
int dx/(cosx+cosa) = 1/sina ln (( 2cos((x-a)/2)cos(a/2) ) / (2cos((x+a)/2)cos(a/2)))+C
int dx/(cosx+cosa) = 1/sina ln ( cos((x-a)/2) / cos((x+a)/2))+C
Further trigonometric semplification is possible, but the answer is getting too long...
