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

The integral of 1/(sec(x)-1) is -cot(x/2)-x+C

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integral of 1/(sec(x)-1)

\int \frac{1}{sec ( x ) - 1} d x

- cot (\frac{x}{2}) - x + C

Solution steps

\int \frac{1}{sec ( x ) - 1} d x

\frac{1}{sec ( x ) - 1} = \frac{sec ( x )}{sec ( x ) - 1} - 1

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

\int \frac{sec ( x )}{sec ( x ) - 1} d x = - cot (\frac{x}{2})

\int 1 d x = x

= - cot (\frac{x}{2}) - x

Add a constant to the solution

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

\int cos^{3} (2 x) d x

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

\frac{d}{d x} (arcsin (9 x) + arccos (9 x))

\int \frac{x ^{3}}{a} - \frac{a}{x ^{3}} d x

\frac{d}{d x} (x^{22})

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