What is the integral of 6/(1-cos(3x)) ?

The integral of 6/(1-cos(3x)) is -2cot((3x)/2)+C

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integral of 6/(1-cos(3x))

\int \frac{6}{1 - cos ( 3 x )} d x

- 2 cot (\frac{3 x}{2}) + C

Solution steps

\int \frac{6}{1 - cos ( 3 x )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 - cos ( 3 x )} d x

Apply u-substitution

= 6 \cdot \int \frac{1}{3 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= 6 \cdot \frac{1}{3} (- \frac{1}{v})

Substitute back

= 6 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )})

Simplify

6 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )}) : - 2 cot (\frac{3 x}{2})

= - 2 cot (\frac{3 x}{2})

Add a constant to the solution

= - 2 cot (\frac{3 x}{2}) + C

What is the integral of 6/(1-cos(3x)) ?
The integral of 6/(1-cos(3x)) is -2cot((3x)/2)+C