What is the integral of 1/(tan(x)(3-2ln(sin(x)))) ?

The integral of 1/(tan(x)(3-2ln(sin(x)))) is -1/2 ln|3-2ln(sin(x))|+C

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integral of 1/(tan(x)(3-2ln(sin(x))))

\int \frac{1}{tan ( x ) ( 3 - 2 ln ( sin ( x ) ) )} d x

- \frac{1}{2} ln ∣ 3 - 2 ln (sin (x)) ∣ + C

Solution steps

\int \frac{1}{tan ( x ) ( 3 - 2 ln ( sin ( x ) ) )} d x

Apply u-substitution

= \int \frac{1}{u ( 3 - 2 ln ( u ) )} d u

Apply u-substitution

= \int - \frac{1}{2 v} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} ln ∣ v ∣

Substitute back

= - \frac{1}{2} ln ∣ 3 - 2 ln (sin (x)) ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 3 - 2 ln (sin (x)) ∣ + C

What is the integral of 1/(tan(x)(3-2ln(sin(x)))) ?
The integral of 1/(tan(x)(3-2ln(sin(x)))) is -1/2 ln|3-2ln(sin(x))|+C