What is the integral of (sin(3t))/(cos(3t)-1) ?

The integral of (sin(3t))/(cos(3t)-1) is -1/3 ln|cos(3t)-1|+C

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integral of (sin(3t))/(cos(3t)-1)

\int \frac{sin ( 3 t )}{cos ( 3 t ) - 1} d t

- \frac{1}{3} ln ∣ cos (3 t) - 1 ∣ + C

Solution steps

\int \frac{sin ( 3 t )}{cos ( 3 t ) - 1} d t

Apply u-substitution

: \int - \frac{1}{3 u} d u

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = cos (3 t) - 1

= - \frac{1}{3} ln ∣ cos (3 t) - 1 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ cos (3 t) - 1 ∣ + C

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What is the integral of (sin(3t))/(cos(3t)-1) ?
The integral of (sin(3t))/(cos(3t)-1) is -1/3 ln|cos(3t)-1|+C