What is the integral of 1/(sin^2(u)) ?

The integral of 1/(sin^2(u)) is -cot(u)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sin^2(u))

\int \frac{1}{sin ^{2} ( u )} d u

- cot (u) + C

Solution steps

\int \frac{1}{sin ^{2} ( u )} d u

Use the common integral:

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

= - cot (u)

Add a constant to the solution

= - cot (u) + C

\frac{d}{d x} ((1 + x)^{k})

\frac{d}{d x} (\frac{5 x ^{2}}{( x ^{2} + x ) ^{3}})

(- 3, - 1), (2, - 2)

h (x) = \frac{1}{x - 2}, (3, 1)

\int \frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}} d x

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