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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 2 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

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