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

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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

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

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

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