# Integrate 1÷(1 + cosec x ) ?

Apr 4, 2018

$\int \frac{1}{1 + \csc x} \mathrm{dx} = \sec x - \tan x + x + c$

#### Explanation:

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

= $\int \frac{\csc x - 1}{\left(\csc x - 1\right) \left(\csc x + 1\right)} \mathrm{dx}$

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

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

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

= $\int \left(\sin \frac{x}{\cos} ^ 2 x - {\sec}^{2} x + 1\right) \mathrm{dx}$

= $\int \left(\sec x \tan x - {\sec}^{2} x + 1\right) \mathrm{dx}$

= $\sec x - \tan x + x + c$

Apr 4, 2018

$\int \frac{1}{1 + \cos e c x}$

$= \int \frac{1}{1 + \frac{1}{\sin} x}$

$= \int \sin \frac{x}{\sin x + 1}$

$= \int \frac{\sin x \left(\sin x - 1\right)}{\left(\sin x + 1\right) \left(\sin x - 1\right)}$

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

$= \int \tan x \sec x \mathrm{dx} - \int {\tan}^{2} x \mathrm{dx}$

$= \int \tan x \sec x \mathrm{dx} - \int \left({\sec}^{2} x - 1\right) \mathrm{dx}$

$= \sec x - \left(\tan x - x\right) + c$

$= x + \sec x - \tan x + c$