To evaluate the integral:

$\int \frac{\cos x}{1 + \sin x} \, dx$

we can use a substitution method.

Step 1: Substitute

Let $u = 1 + \sin x$. Then $du = \cos x \, dx$.

This changes the integral to:

$\int \frac{\cos x}{1 + \sin x} \, dx = \int \frac{1}{u} \, du$

Step 2: Integrate

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln |u|$:

$\int \frac{1}{u} \, du = \ln |u| + C$

Step 3: Substitute Back

Since $u = 1 + \sin x$, we substitute back to get:

$\ln |1 + \sin x| + C$

Final Answer

$\int \frac{\cos x}{1 + \sin x} \, dx = \ln |1 + \sin x| + C$

Would you like further explanation, or do you have any questions?

1. How does substitution help in evaluating integrals?
2. Can this technique be applied to other trigonometric integrals?
3. What are common substitution choices in trigonometric integrals?
4. How can we confirm that our answer is correct?
5. What other techniques can simplify integrals involving trigonometric functions?

Tip: In trigonometric integrals, look for identities or substitutions that simplify the expression into basic forms.