What is the integral of (18)/(sec(2x)-1) ?

The integral of (18)/(sec(2x)-1) is 9(-2x-cot(x))+C

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integral of (18)/(sec(2x)-1)

\int \frac{18}{sec ( 2 x ) - 1} d x

9 (- 2 x - cot (x)) + C

Solution steps

\int \frac{18}{sec ( 2 x ) - 1} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{sec ( 2 x ) - 1} d x

Apply u-substitution

: \int \frac{1}{2 ( sec ( u ) - 1 )} d u

= 18 \cdot \int \frac{1}{2 ( sec ( u ) - 1 )} d u

Take the constant out:

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

= 18 \cdot \frac{1}{2} \cdot \int \frac{1}{sec ( u ) - 1} d u

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

= 18 \cdot \frac{1}{2} \cdot \int \frac{sec ( u )}{sec ( u ) - 1} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 18 \cdot \frac{1}{2} (\int \frac{sec ( u )}{sec ( u ) - 1} d u - \int 1 d u)

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

\int 1 d u = u

= 18 \cdot \frac{1}{2} (- cot (\frac{u}{2}) - u)

Substitute back

u = 2 x

= 18 \cdot \frac{1}{2} (- cot (\frac{2 x}{2}) - 2 x)

Simplify

18 \cdot \frac{1}{2} (- cot (\frac{2 x}{2}) - 2 x) : 9 (- 2 x - cot (x))

= 9 (- 2 x - cot (x))

Add a constant to the solution

= 9 (- 2 x - cot (x)) + C

\int \frac{1}{e ^{x} + e ^{- x} + 2} d x

8 y^{^{''}} + y^{^{'}} = 0

\int w^{3} e^{w^{2}} d w

\frac{e ^{\frac{x}{2}}}{x + 1}

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{4}})

What is the integral of (18)/(sec(2x)-1) ?
The integral of (18)/(sec(2x)-1) is 9(-2x-cot(x))+C