Question

Evaluate: $\int \frac{1}{1+\sin \theta }d\theta$

Answer 1

$I=\int \frac{1}{1+\sin \theta }d\theta$

$I=\int \frac{1-\sin \theta }{(1+\sin \theta )(1-\sin \theta )}d\theta =\int \frac{1-\sin \theta }{1-{\sin }^2\theta }d\theta$

$I=\int \frac{1-\sin \theta }{{\cos }^2\theta }d\theta =\int ({\sec }^2\theta -\sec \theta \tan \theta )d\theta$

$I=\tan \theta -\sec \theta +c$.