What is the integral of (12)/(1-cos(3x)) ?

The integral of (12)/(1-cos(3x)) is -4cot((3x)/2)+C

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integral of (12)/(1-cos(3x))

\int \frac{12}{1 - cos ( 3 x )} d x

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

Solution steps

\int \frac{12}{1 - cos ( 3 x )} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{1 - cos ( 3 x )} d x

Apply u-substitution

: \int \frac{1}{3 ( 1 - cos ( u ) )} d u

= 12 \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

= 12 \cdot \frac{1}{3} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 12 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 12 \cdot \frac{1}{3} (- \frac{1}{v})

Substitute back

= 12 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )})

Simplify

12 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )}) : - 4 cot (\frac{3 x}{2})

= - 4 cot (\frac{3 x}{2})

Add a constant to the solution

= - 4 cot (\frac{3 x}{2}) + C

\frac{\partial}{\partial y} (4 x^{3} - 4 x)

3^{2 x + 1}

\int \frac{1}{7 x - 2} d x

x \to 3 - lim (x - 3)

x \to 1 lim (\frac{x + 9}{x + 1})

What is the integral of (12)/(1-cos(3x)) ?
The integral of (12)/(1-cos(3x)) is -4cot((3x)/2)+C