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

The integral of (sec(x)tan(x))/(sec(x)-1) is ln|sec(x)-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ sec (x) - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = sec (x) - 1

= ln ∣ sec (x) - 1 ∣

Add a constant to the solution

= ln ∣ sec (x) - 1 ∣ + C

(x + 2) (2 x^{2} - 3)

\int x e^{k x} d x

s + 1 - 2.82 j

\int 3 x^{2} - 18 d x

x \to 0 lim (x^{2} cos (20 π x))

What is the integral of (sec(x)tan(x))/(sec(x)-1) ?
The integral of (sec(x)tan(x))/(sec(x)-1) is ln|sec(x)-1|+C