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

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

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

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

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

Solution steps

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

Factor

1 - sec (x) : - (- 1 + sec (x))

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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

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