∫cos(x)/(1+sin(x)) dx =
∫d[sin(x)]/(1+sin(x))=
Ln[1 + sin(x)]
So, the integral from zero to pi/2 is
Ln(2) - Ln(1) = Ln(2)
i got the wrong answer. the correct answer is ln(2).
here is problem:
∫cosx/(1+sinx) dx from 0 to pi/2
2 answers
thanks