What is the integral of (16)/(sec(2x)-1) ?

The integral of (16)/(sec(2x)-1) is 8(-2x-cot(x))+C

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integral of (16)/(sec(2x)-1)

\int \frac{16}{sec ( 2 x ) - 1} d x

8 (- 2 x - cot (x)) + C

Solution steps

\int \frac{16}{sec ( 2 x ) - 1} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{sec ( 2 x ) - 1} d x

Apply u-substitution

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

= 16 \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

= 16 \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

= 16 \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

= 16 \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

= 16 \cdot \frac{1}{2} (- cot (\frac{u}{2}) - u)

Substitute back

u = 2 x

= 16 \cdot \frac{1}{2} (- cot (\frac{2 x}{2}) - 2 x)

Simplify

16 \cdot \frac{1}{2} (- cot (\frac{2 x}{2}) - 2 x) : 8 (- 2 x - cot (x))

= 8 (- 2 x - cot (x))

Add a constant to the solution

= 8 (- 2 x - cot (x)) + C

\frac{d}{d x} (2 e^{2 x} + 3 e^{3 x})

x \to a lim ([f (x)]^{g (x)})

t \to 0 lim (\frac{3}{t} - \frac{3}{e ^{t} - 1})

y^{^{'}} + y = t e^{- t} + 1

\int \frac{1}{12} d x

What is the integral of (16)/(sec(2x)-1) ?
The integral of (16)/(sec(2x)-1) is 8(-2x-cot(x))+C