What is the integral of (18)/(1-cos(3x)) ?

The integral of (18)/(1-cos(3x)) is -6cot((3x)/2)+C

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integral of (18)/(1-cos(3x))

\int \frac{18}{1 - cos ( 3 x )} d x

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

Solution steps

\int \frac{18}{1 - cos ( 3 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 ( 3 x )} d x

Apply u-substitution

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

= 18 \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

= 18 \cdot \frac{1}{3} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 18 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 18 \cdot \frac{1}{3} (- \frac{1}{v})

Substitute back

= 18 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )})

Simplify

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

= - 6 cot (\frac{3 x}{2})

Add a constant to the solution

= - 6 cot (\frac{3 x}{2}) + C

- 7 x^{3} e^{x} - 21 x^{2} e^{x}

x \to 2 + lim ((x - 2)^{2})

3 x^{2} + 6 x + 4

\frac{d}{d x} (\frac{2 x}{4 x - 3})

What is the integral of (18)/(1-cos(3x)) ?
The integral of (18)/(1-cos(3x)) is -6cot((3x)/2)+C