The integral to be determined is `int sin x/(cos x + sin x) dx`

Using a little lateral thinking makes determining the given integral easy. The derivative of sin x + cos x is cos x - sin x.

`int sin x/(cos x + sin x) dx`

can be written as: `int (1/2)*(sin x + sin x + cos x - cos x)/(cos x + sin x) dx`

=> `(1/2)*int ((sin x + cos x) - (cos x - sin x))/(cos x + sin x) dx`

=> `(1/2)*int (sin x + cos x)/(cos x + sin x) dx - (1/2)*int (cos x - sin x)/(cos x + sin x) dx`

=> `(1/2)* int 1 dx - (1/2)*int (cos x - sin x)/(cos x + sin x) dx`

let `cos x + sin x = y`

`dy = (cos x - sin x) dx`

=> `(1/2)* int 1 dx - (1/2)*int (1/y) dy`

=> `(1/2)*x - (1/2)*ln y + C`

substitute y = `sin x + cos x`

=> `(1/2)x - (1/2)*ln(cos x + sin x) + C`

**The integral `int sin x/(cos x + sin x) dx` = `(x - ln (cos x + sin x))/2 + C` **