# int_0^(pi/2) 1/(1+sin(theta)+cos(theta)) d theta Find or evaluate the integral

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