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

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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`

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