To find the indefinite integral [tex]\(\int \frac{\cos t}{1+\sin t} \, dt\)[/tex], let's proceed with the following steps:
1. Substitution:
Let's use the substitution [tex]\(u = 1 + \sin t\)[/tex]. Then we need to find the differential [tex]\(du\)[/tex].
Recall that the derivative of [tex]\(\sin t\)[/tex] with respect to [tex]\(t\)[/tex] is [tex]\(\cos t\)[/tex]. Hence, we have:
[tex]\[
du = \cos t \, dt
\][/tex]
This implies that [tex]\( \cos t \, dt = du \)[/tex].
2. Rewrite the integral:
Using the substitution, the integral [tex]\(\int \frac{\cos t}{1+\sin t} \, dt\)[/tex] transforms to:
[tex]\[
\int \frac{du}{u}
\][/tex]
Here we made the substitution [tex]\(\cos t \, dt = du\)[/tex] and [tex]\(u = 1 + \sin t\)[/tex].
3. Integrate:
The integral [tex]\(\int \frac{du}{u}\)[/tex] is a standard integral and is given by:
[tex]\[
\int \frac{du}{u} = \ln |u| + C
\][/tex]
4. Substitute back [tex]\(u\)[/tex]:
Recall that [tex]\(u = 1 + \sin t\)[/tex]. Substitute back into the expression for the integral:
[tex]\[
\ln |1 + \sin t| + C
\][/tex]
Therefore, the indefinite integral [tex]\(\int \frac{\cos t}{1+\sin t} \, dt\)[/tex] evaluates to:
[tex]\[
\boxed{\ln |1 + \sin t| + C}
\][/tex]
