What is the integral of 1/(16sin^2(x)) ?

The integral of 1/(16sin^2(x)) is -1/16 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/(16sin^2(x))

\int \frac{1}{16 sin ^{2} ( x )} d x

- \frac{1}{16} cot (x) + C

Solution steps

\int \frac{1}{16 sin ^{2} ( x )} d x

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{sin ^{2} ( x )} d x

Use the common integral:

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

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

Simplify

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

Add a constant to the solution

= - \frac{1}{16} cot (x) + C

81 y^{^{''}} + 18 y^{^{'}} + y = 0

(1.2) (2.4)

\frac{d}{d x} (\frac{4 x ^{3} + 3 x ^{2}}{x})

sin (x), \frac{1}{2}

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