What is the integral of a/(1-cos(bx)) ?

The integral of a/(1-cos(bx)) is -a/(btan(\frac{bx){2})}+C

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integral of a/(1-cos(bx))

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

- \frac{a}{b tan ( \frac{b x}{2} )} + C

Solution steps

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

Take the constant out:

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

= a \cdot \int \frac{1}{1 - cos ( b x )} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= a \frac{1}{b} (- \frac{1}{v})

Substitute back

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

Simplify

a \frac{1}{b} (- \frac{1}{tan ( \frac{b x}{2} )}) : - \frac{a}{b tan ( \frac{b x}{2} )}

= - \frac{a}{b tan ( \frac{b x}{2} )}

Add a constant to the solution

= - \frac{a}{b tan ( \frac{b x}{2} )} + C

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What is the integral of a/(1-cos(bx)) ?
The integral of a/(1-cos(bx)) is -a/(btan(\frac{bx){2})}+C