What is the integral of 9/(sec(x)-1) ?

The integral of 9/(sec(x)-1) is 9(-cot(x/2)-x)+C

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integral of 9/(sec(x)-1)

\int \frac{9}{sec ( x ) - 1} d x

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

Solution steps

\int \frac{9}{sec ( x ) - 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{sec ( x ) - 1} d x

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

= 9 \cdot \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

= 9 (\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

= 9 (- cot (\frac{x}{2}) - x)

Add a constant to the solution

= 9 (- cot (\frac{x}{2}) - x) + C

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What is the integral of 9/(sec(x)-1) ?
The integral of 9/(sec(x)-1) is 9(-cot(x/2)-x)+C