Question

Integral of 1÷(1+sin x) dx
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Answer 1

`\tan \: x - \sec \: x + c`

Explanation:

`\int1 / (1 + \sin \: x) \: dx \\ \\ = \int (1)/(1 + \sin \: x) dx \\ \\ = \int (1)/((1 + \sin \: x)) * ((1 - \sin \: x))/((1 - \sin \: x)) dx \\ \\ = \int (1 - \sin \: x)/(1 - \sin^(2) \: x) dx \\ \\ = \int (1 - \sin \: x)/( \cos^(2) \: x) dx \\ \\ = \int \bigg( (1)/( \cos^(2) \: x) - (\sin \: x)/( \cos^(2) \: x) \bigg) dx \\ \\ = \int( { \sec}^(2) x - \sec \: x. \tan \: x)dx \\ \\ \purple{ \bold{ = \tan \: x - \sec \: x + c}}`