What is the integral of 7/(sec(x)-1) ?

The integral of 7/(sec(x)-1) is 7(-cot(x/2)-x)+C

integral of 7/(sec(x)-1)

\int \frac{7}{sec ( x ) - 1} d x

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

Solution steps

\int \frac{7}{sec ( x ) - 1} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{sec ( x ) - 1} d x

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

= 7 \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

= 7 (\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

= 7 (- cot (\frac{x}{2}) - x)

Add a constant to the solution

= 7 (- cot (\frac{x}{2}) - x) + C

\int \frac{e ^{2 x}}{e ^{5 - x}} d x

x \to \infty lim (67)

\int_{- 2}^{x} \frac{x + 2}{3} d x

d er i v a t i v e \frac{3}{x ^{9}}

x \to \infty lim ((x + 1)^{x + 2})

What is the integral of 7/(sec(x)-1) ?
The integral of 7/(sec(x)-1) is 7(-cot(x/2)-x)+C