What is the integral of 2/(cos(3x)-1) ?

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

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integral of 2/(cos(3x)-1)

\int \frac{2}{cos ( 3 x ) - 1} d x

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

Solution steps

\int \frac{2}{cos ( 3 x ) - 1} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{cos ( 3 x ) - 1} d x

Factor

cos (3 x) - 1 : - (- cos (3 x) + 1)

= 2 \cdot \int \frac{1}{- ( - cos ( 3 x ) + 1 )} d x

Take the constant out:

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

= 2 (- \int \frac{1}{- cos ( 3 x ) + 1} d x)

Apply u-substitution

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

= 2 (- \int \frac{1}{3 ( - cos ( u ) + 1 )} d u)

Take the constant out:

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

= 2 (- \frac{1}{3} \cdot \int \frac{1}{- cos ( u ) + 1} d u)

Apply u-substitution

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

= 2 (- \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v)

Apply the Power Rule

: - \frac{1}{v}

= 2 (- \frac{1}{3} (- \frac{1}{v}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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