\( \int _{}^{}\frac{1}{1+\sin (x)} d x \)

To solve the integral \( \int \frac{1}{1+\sin(x)} \, dx \), we can simplify the expression by multiplying the numerator and denominator by \( 1 - \sin(x) \), which gives us: \[ \int \frac{1 - \sin(x)}{(1+\sin(x))(1-\sin(x))} \, dx = \int \frac{1 - \sin(x)}{\cos^2(x)} \, dx \] This results in: \[ \int \sec^2(x) \, dx - \int \frac{\sin(x)}{\cos^2(x)} \, dx \] The first integral is \( \tan(x) + C \), while the second integral can be solved by recognizing it's the derivative of \( -\frac{1}{\cos(x)} \) or simply: \[ \int \tan(x) \sec(x) \, dx = -\sec(x) + C \] Combining the results yields: \[ \tan(x) - \sec(x) + C \] So, the final answer to your integral is: \[ \tan(x) - \sec(x) + C \] Enjoy your exploration into the realm of calculus!