What is the integral of 1/(sin^2(w)(1-4cot(w))) ?

The integral of 1/(sin^2(w)(1-4cot(w))) is 1/4 ln|1-4cot(w)|+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(w)(1-4cot(w)))

\int \frac{1}{sin ^{2} ( w ) ( 1 - 4 cot ( w ) )} d w

\frac{1}{4} ln ∣ 1 - 4 cot (w) ∣ + C

Solution steps

\int \frac{1}{sin ^{2} ( w ) ( 1 - 4 cot ( w ) )} d w

Apply u-substitution

: \int - \frac{1}{1 - 4 u} d u

= \int - \frac{1}{1 - 4 u} d u

Take the constant out:

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

= - \int \frac{1}{1 - 4 u} d u

Apply u-substitution

: \int - \frac{1}{4 v} d v

= - \int - \frac{1}{4 v} d v

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - (- \frac{1}{4} ln ∣ v ∣)

Substitute back

= - (- \frac{1}{4} ln ∣ 1 - 4 cot (w) ∣)

Simplify

= \frac{1}{4} ln ∣ 1 - 4 cot (w) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 1 - 4 cot (w) ∣ + C

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

\int \frac{2}{x} - 1 d x

\int \frac{( ( ln ( x ) ) ^{4} )}{x} d x

\int 2 x (x^{2} + 6)^{5} d x

\int \frac{1}{2 x + x ^{2}} d x

What is the integral of 1/(sin^2(w)(1-4cot(w))) ?
The integral of 1/(sin^2(w)(1-4cot(w))) is 1/4 ln|1-4cot(w)|+C