How to integrate 1/(1+sinx)?

I'm currently learning Trigonometric Functions and need help with solving two questions: $\int$ $\frac{1}{1 + \sin x}$ $\mathrm{dx}$ = ? $\int$ $\frac{1}{1 + \cos 2 x}$ $\mathrm{dx}$ = ? Thanks!

Oct 2, 2017

$\int \frac{1}{1 + \sin x} \mathrm{dx} = \tan x - \sec x$ and $\int \frac{1}{1 + \cos 2 x} \mathrm{dx} = \frac{1}{2} \tan x$

Explanation:

$\int \frac{1}{1 + \sin x} \mathrm{dx}$

= $\int \frac{1 - \sin x}{1 - {\sin}^{2} x} \mathrm{dx}$

= $\int \frac{1 - \sin x}{\cos} ^ 2 x \mathrm{dx}$

= $\int \left({\sec}^{2} x - \tan x \sec x\right) \mathrm{dx}$

= $\tan x - \sec x$

$\int \frac{1}{1 + \cos 2 x} \mathrm{dx}$

= $\int \frac{1}{1 + 2 {\cos}^{2} x - 1} \mathrm{dx}$

= $\int \frac{1}{2 {\cos}^{2} x} \mathrm{dx}$

= $\frac{1}{2} \int {\sec}^{2} x \mathrm{dx}$

= $\frac{1}{2} \tan x$