What is the integral of (12)/(1-cos(6x)) ?

The integral of (12)/(1-cos(6x)) is -2cot(3x)+C

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integral of (12)/(1-cos(6x))

\int \frac{12}{1 - cos ( 6 x )} d x

- 2 cot (3 x) + C

Solution steps

\int \frac{12}{1 - cos ( 6 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 ( 6 x )} d x

Apply u-substitution

: \int \frac{1}{6 ( 1 - cos ( u ) )} d u

= 12 \cdot \int \frac{1}{6 ( 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}{6} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 12 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 12 \cdot \frac{1}{6} (- \frac{1}{v})

Substitute back

= 12 \cdot \frac{1}{6} (- \frac{1}{tan ( \frac{6 x}{2} )})

Simplify

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

= - 2 cot (3 x)

Add a constant to the solution

= - 2 cot (3 x) + C

x^{3} + y^{3} = 126

\frac{d}{d x} (3 x^{2} (2 x + 1)^{5})

n = 1 \sum \infty (\frac{n}{n + 5})^{n}

\frac{e ^{- 5 s}}{s ^{2}}

\frac{\partial}{\partial z} (y^{2} z^{3})

What is the integral of (12)/(1-cos(6x)) ?
The integral of (12)/(1-cos(6x)) is -2cot(3x)+C