Asked by Vicky
i got the wrong answer. the correct answer is ln(2).
here is problem:
∫cosx/(1+sinx) dx from 0 to pi/2
here is problem:
∫cosx/(1+sinx) dx from 0 to pi/2
Answers
Answered by
Count Iblis
∫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)
∫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)
Answered by
Vicky
thanks
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