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

The integral of 1/(tan(x)-1) is 1/2 ln|tan^2(x/2)+2tan(x/2)-1|-1/2 ln(sec^2(x/2))-arctan(tan(x/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 - \frac{1}{2} ln (sec^{2} (\frac{x}{2})) - arctan (tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{2 ( 1 - u ^{2} )}{( 2 u - 1 + u ^{2} ) ( 1 + u ^{2} )} d u

Take the constant out:

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

= 2 \cdot \int \frac{1 - u ^{2}}{( 2 u - 1 + u ^{2} ) ( 1 + u ^{2} )} d u

Take the partial fraction of

\frac{1 - u ^{2}}{( 2 u - 1 + u ^{2} ) ( 1 + u ^{2} )} : \frac{u + 1}{2 ( u ^{2} + 2 u - 1 )} + \frac{- u - 1}{2 ( u ^{2} + 1 )}

= 2 \cdot \int \frac{u + 1}{2 ( u ^{2} + 2 u - 1 )} + \frac{- u - 1}{2 ( u ^{2} + 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

= 2 (\int \frac{u + 1}{2 ( u ^{2} + 2 u - 1 )} d u + \int \frac{- u - 1}{2 ( u ^{2} + 1 )} d u)

\int \frac{u + 1}{2 ( u ^{2} + 2 u - 1 )} d u = \frac{1}{4} ln u^{2} + 2 u - 1

\int \frac{- u - 1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u))

= 2 (\frac{1}{4} ln u^{2} + 2 u - 1 + \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u)))

Substitute back

u = tan (\frac{x}{2})

= 2 (\frac{1}{4} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 + \frac{1}{2} (- \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 - arctan (tan (\frac{x}{2}))))

Simplify

2 (\frac{1}{4} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 + \frac{1}{2} (- \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 - arctan (tan (\frac{x}{2})))) : \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 - \frac{1}{2} ln (sec^{2} (\frac{x}{2})) - arctan (tan (\frac{x}{2}))

= \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 - \frac{1}{2} ln (sec^{2} (\frac{x}{2})) - arctan (tan (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 - \frac{1}{2} ln (sec^{2} (\frac{x}{2})) - arctan (tan (\frac{x}{2})) + C

\int_{- 5}^{6} x d x

f (x) = (x^{2} - 1) (1 - e^{x})

\int 4 sin^{2} (x co) s^{3} x d x

\int 17 d x

\int_{- 6}^{0} (2 + 36 - x^{2}) d x

What is the integral of 1/(tan(x)-1) ?
The integral of 1/(tan(x)-1) is 1/2 ln|tan^2(x/2)+2tan(x/2)-1|-1/2 ln(sec^2(x/2))-arctan(tan(x/2))+C