Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 55

To evaluate the integral problem: int_0^(pi/2) 1/(1+sin(theta)+cos(theta)) d theta , we may apply Weierstrass substitution or tangent half-angle substitution .
This helps to determine the indefinite integral of a rational function in terms of sine and cosine. We let:
u = tan(theta/2)
sin(theta) = (2u)/(1+u^2)
cos(theta) =(1-u^2)/(1+u^2)
d theta=(2 du)/(1+u^2)
Plug-in the values to express the integral problem in terms variable "u'.
int 1/(1+sin(theta)+cos(theta)) d theta=int 1/(1+(2u)/(1+u^2)+(1-u^2)/(1+u^2))*(2 du)/(1+u^2)
=int 1/(((1+u^2)/(1+u^2)+(2u)/(1+u^2)+(1-u^2)/(1+u^2)))*(2 du)/(1+u^2)
=int 1/(((1+u^2+ 2u +1-u^2)/(1+u^2)))*(2 du)/(1+u^2)
=int 1/(((2 +2u)/(1+u^2)))*(2 du)/(1+u^2)
=int 1 *(1+u^2)/ (2 +2u)*(2 du)/(1+u^2)
=int (2 du)/ (2 +2u)
=int (2 du)/ (2(1 +u))
=int (du)/(1+u)

From the table of indefinite integration table, we follow the integral formula for rational function as:
int (dx)/(ax+b)=1/aln(ax+b)
By comparing "ax+b " with "1+u or 1u +1 ", the corresponding values are: a=1 and b=1 . Then, the integral becomes:
int (du)/(1+u)=1/1ln(1u+1)
=ln(u+1)
Plug-in u =tan(x/2) on ln(u+1) , we get:
int_0^(pi/2) 1/(1+sin(theta)+cos(theta)) d theta=ln(tan(x/2)+1)|_0^(pi/2)
Apply the definite integral formula: F(x)|_a^b= F(b)-F(a) .
ln(tan(x/2)+1)|_0^(pi/2)=ln(tan(((pi/2))/2)+1)-ln(tan(0/2)+1)
=ln(tan(pi/4)+1)-ln(tan(0)+1)
=ln(1+1)-ln(0+1)
=ln(2)-ln(1)
= ln(2/1)
=ln(2) or 0.693