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

The integral of (tan(x))/(1-cos(x)) is ln(tan^2(x/2))-ln|tan^2(x/2)-1|+C

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

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

ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ( - u ^{2} + 1 )} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 v ( - v + 1 )} d v

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{v ( - v + 1 )} : \frac{1}{v} - \frac{1}{v - 1}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v} - \frac{1}{v - 1} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= 2 \cdot \frac{1}{2} (ln ∣ v ∣ - ln ∣ v - 1 ∣)

Substitute back

= 2 \cdot \frac{1}{2} (ln tan^{2} (\frac{x}{2}) - ln tan^{2} (\frac{x}{2}) - 1)

Simplify

2 \cdot \frac{1}{2} (ln tan^{2} (\frac{x}{2}) - ln tan^{2} (\frac{x}{2}) - 1) : ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1

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

Add a constant to the solution

= ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1 + C

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What is the integral of (tan(x))/(1-cos(x)) ?
The integral of (tan(x))/(1-cos(x)) is ln(tan^2(x/2))-ln|tan^2(x/2)-1|+C