What is the integral of (24)/(1-cos(6x)) ?

The integral of (24)/(1-cos(6x)) is -4cot(3x)+C

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integral of (24)/(1-cos(6x))

\int \frac{24}{1 - cos ( 6 x )} d x

- 4 cot (3 x) + C

Solution steps

\int \frac{24}{1 - cos ( 6 x )} d x

Take the constant out:

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

= 24 \cdot \int \frac{1}{1 - cos ( 6 x )} d x

Apply u-substitution

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

= 24 \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

= 24 \cdot \frac{1}{6} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 24 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 24 \cdot \frac{1}{6} (- \frac{1}{v})

Substitute back

= 24 \cdot \frac{1}{6} (- \frac{1}{tan ( \frac{6 x}{2} )})

Simplify

24 \cdot \frac{1}{6} (- \frac{1}{tan ( \frac{6 x}{2} )}) : - 4 cot (3 x)

= - 4 cot (3 x)

Add a constant to the solution

= - 4 cot (3 x) + C

sin (x) + cos (x)

\int_{0}^{4} (1 + 2 x)^{2} d x

(1, 2), (3, 16)

n = 0 \sum \infty (\frac{n}{n + 10})^{n}

\int ((\frac{- x ^{2} + 8}{2}) - (\frac{x ^{2}}{2})) d x

What is the integral of (24)/(1-cos(6x)) ?
The integral of (24)/(1-cos(6x)) is -4cot(3x)+C