What is the integral of (1/(1-sin(x))-1/(1+sin(x))) ?

The integral of (1/(1-sin(x))-1/(1+sin(x))) is -2/(tan(x/2)-1)+2/(tan(x/2)+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1/(1-sin(x))-1/(1+sin(x)))

\int (\frac{1}{1 - sin ( x )} - \frac{1}{1 + sin ( x )}) d x

- \frac{2}{tan ( \frac{x}{2} ) - 1} + \frac{2}{tan ( \frac{x}{2} ) + 1} + C

Solution steps

\int \frac{1}{1 - sin ( x )} - \frac{1}{1 + sin ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{1 - sin ( x )} d x - \int \frac{1}{1 + sin ( x )} d x

\int \frac{1}{1 - sin ( x )} d x = - \frac{2}{tan ( \frac{x}{2} ) - 1}

\int \frac{1}{1 + sin ( x )} d x = - \frac{2}{tan ( \frac{x}{2} ) + 1}

= - \frac{2}{tan ( \frac{x}{2} ) - 1} - (- \frac{2}{tan ( \frac{x}{2} ) + 1})

Simplify

= - \frac{2}{tan ( \frac{x}{2} ) - 1} + \frac{2}{tan ( \frac{x}{2} ) + 1}

Add a constant to the solution

= - \frac{2}{tan ( \frac{x}{2} ) - 1} + \frac{2}{tan ( \frac{x}{2} ) + 1} + C

What is the integral of (1/(1-sin(x))-1/(1+sin(x))) ?
The integral of (1/(1-sin(x))-1/(1+sin(x))) is -2/(tan(x/2)-1)+2/(tan(x/2)+1)+C