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

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

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

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

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

Solution steps

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

Factor

1 - sec (t) : - (- 1 + sec (t))

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

Take the constant out:

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

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

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

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

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 ( t )}{- 1 + sec ( t )} d t - \int 1 d t)

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

\int 1 d t = t

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

Simplify

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

Add a constant to the solution

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

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