What is the integral of (18)/(1-cos(9x)) ?

The integral of (18)/(1-cos(9x)) is -2cot((9x)/2)+C

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integral of (18)/(1-cos(9x))

\int \frac{18}{1 - cos ( 9 x )} d x

- 2 cot (\frac{9 x}{2}) + C

Solution steps

\int \frac{18}{1 - cos ( 9 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 ( 9 x )} d x

Apply u-substitution

: \int \frac{1}{9 ( 1 - cos ( u ) )} d u

= 18 \cdot \int \frac{1}{9 ( 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}{9} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 18 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 18 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 18 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )})

Simplify

18 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )}) : - 2 cot (\frac{9 x}{2})

= - 2 cot (\frac{9 x}{2})

Add a constant to the solution

= - 2 cot (\frac{9 x}{2}) + C

\frac{2 x 3 x - 1}{5 x}

\int cot^{5} (2 x) d x

\int \frac{4 x ^{2} - 13 x + 12}{x ( x - 2 ) ^{2}} d x

\int \frac{3 x ^{3}}{x ^{2} + 16} d x

\frac{d}{d x} (\frac{e ^{x}}{cos ( 3 x )})

What is the integral of (18)/(1-cos(9x)) ?
The integral of (18)/(1-cos(9x)) is -2cot((9x)/2)+C