What is the integral of 6/(1-cos(3x)) ?

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

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integral of 6/(1-cos(3x))

\int \frac{6}{1 - cos ( 3 x )} d x

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

Solution steps

\int \frac{6}{1 - cos ( 3 x )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 - cos ( 3 x )} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 6 \cdot \frac{1}{3} (- \frac{1}{v})

Substitute back

= 6 \cdot \frac{1}{3} (- \frac{1}{tan ( \frac{3 x}{2} )})

Simplify

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

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

Add a constant to the solution

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

\frac{( x ^{2} + 4 x + 3 )}{x}

\int x 1 0^{x} d x

(\frac{9 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

x^{2} - 2 x + 4, 2

\int_{0}^{1} \frac{1}{1 - x} d x

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