What is the integral of 1/(1-cos(4x)) ?

The integral of 1/(1-cos(4x)) is -1/4 cot(2x)+C

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integral of 1/(1-cos(4x))

\int \frac{1}{1 - cos ( 4 x )} d x

- \frac{1}{4} cot (2 x) + C

Solution steps

\int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( u ) )} d u

= \int \frac{1}{4 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{4} (- \frac{1}{v})

Substitute back

= \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

\frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - \frac{1}{4} cot (2 x)

= - \frac{1}{4} cot (2 x)

Add a constant to the solution

= - \frac{1}{4} cot (2 x) + C

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What is the integral of 1/(1-cos(4x)) ?
The integral of 1/(1-cos(4x)) is -1/4 cot(2x)+C