What is the integral of 1/(tan^2(x)) ?

The integral of 1/(tan^2(x)) is -x-cot(x)+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^2(x))

\int \frac{1}{tan ^{2} ( x )} d x

- x - cot (x) + C

Solution steps

\int \frac{1}{tan ^{2} ( x )} d x

Rewrite using trig identities

= \int - 1 + csc^{2} (x) 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

= - \int 1 d x + \int csc^{2} (x) d x

\int 1 d x = x

\int csc^{2} (x) d x = - cot (x)

= - x - cot (x)

Add a constant to the solution

= - x - cot (x) + C

\int 1 \frac{1}{t} d t

x^{11} arccos (x)

x^{3} - 8 x^{2} + 18 x - 5, x + 5

\frac{d}{d x} (\frac{3 x + 2}{x ^{2} + 1})

g (x) = e^{x^{5}}, (- 1, \frac{1}{e})

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