What is the integral of (18)/(1-cos(6x)) ?

The integral of (18)/(1-cos(6x)) is -3cot(3x)+C

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integral of (18)/(1-cos(6x))

\int \frac{18}{1 - cos ( 6 x )} d x

- 3 cot (3 x) + C

Solution steps

\int \frac{18}{1 - cos ( 6 x )} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{1 - cos ( 6 x )} d x

Apply u-substitution

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

= 18 \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

= 18 \cdot \frac{1}{6} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 18 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 18 \cdot \frac{1}{6} (- \frac{1}{v})

Substitute back

= 18 \cdot \frac{1}{6} (- \frac{1}{tan ( \frac{6 x}{2} )})

Simplify

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

= - 3 cot (3 x)

Add a constant to the solution

= - 3 cot (3 x) + C

sin (t - \frac{( 2 \cdot π )}{3})

(- 3 x + 2)^{4} (- 2 x + 3)^{3}

x \to 0 lim (\frac{2 tan ( \frac{x}{2} )}{x})

\int tan^{5} (6 x) sec^{2} (6 x) d x

\frac{d}{d x} (\frac{1}{9 x ^{\frac{8}{9}}})

What is the integral of (18)/(1-cos(6x)) ?
The integral of (18)/(1-cos(6x)) is -3cot(3x)+C