Answer:
[tex] \tan \: x - \sec \: x + c[/tex]
Step-by-step explanation:
[tex] \int1 \div (1 + \sin \: x) \: dx \\ \\ = \int \frac{1}{1 + \sin \: x} dx \\ \\ = \int \frac{1}{(1 + \sin \: x)} \times \frac{(1 - \sin \: x)}{(1 - \sin \: x)} dx \\ \\ = \int \frac{1 - \sin \: x}{1 - \sin^{2} \: x} dx \\ \\ = \int \frac{1 - \sin \: x}{ \cos^{2} \: x} dx \\ \\ = \int \bigg( \frac{1}{ \cos^{2} \: x} - \frac{\sin \: x}{ \cos^{2} \: x} \bigg) dx \\ \\ = \int( { \sec}^{2} x - \sec \: x. \tan \: x)dx \\ \\ \purple{ \bold{ = \tan \: x - \sec \: x + c}}[/tex]
