What is the integral of (sin(ax))/(1-cos(ax)) ?

The integral of (sin(ax))/(1-cos(ax)) is 1/a ln|1-cos(ax)|+C

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integral of (sin(ax))/(1-cos(ax))

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

\frac{1}{a} ln ∣ 1 - cos (a x) ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{a u} d u

= \int \frac{1}{a u} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{a} ln ∣ u ∣

Substitute back

u = 1 - cos (a x)

= \frac{1}{a} ln ∣ 1 - cos (a x) ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ 1 - cos (a x) ∣ + C

x \to 0 lim (\frac{5}{x ^{2} - x})

\frac{d}{d x} (\frac{32 x}{( x ^{2} + 3 ) ^{2}})

4 t^{- \frac{3}{8}}

\int \frac{( x + 7 ) ^{7}}{2 x} d x

\int \frac{2}{( x ^{2} + 1 ) ^{2}} d x

What is the integral of (sin(ax))/(1-cos(ax)) ?
The integral of (sin(ax))/(1-cos(ax)) is 1/a ln|1-cos(ax)|+C