i got the wrong answer. the correct answer is ln(2).
here is problem:
∫cosx/(1+sinx) dx from 0 to pi/2
∫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)
thanks
To evaluate the integral ∫cosx/(1+sinx) dx from 0 to π/2, you can use the substitution method.
Let's start by making the substitution u = 1+sinx. To do this, we need to find the derivative of u with respect to x.
Differentiating both sides of u = 1+sinx with respect to x, we get du/dx = cosx.
Now, rewrite the integral with respect to u:
∫cosx/(1+sinx) dx = ∫du/u
We can simplify further by canceling out the common term of cosx and dx. Now, the integral becomes:
∫du/u
Integrating this expression gives:
ln|u| + C
where C is the constant of integration.
Now, substitute u back in:
ln|1+sinx| + C
To evaluate the definite integral from 0 to π/2, substitute the upper and lower limits of integration:
ln|1+sin(π/2)| - ln|1+sin(0)|
ln|1+1| - ln|1+0|
ln(2) - ln(1)
ln(2) is the correct answer for the definite integral of cosx/(1+sinx) dx from 0 to π/2.