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

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

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

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

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

Solution steps

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

Factor

cos (x) - sin (x) - 1 : - (- cos (x) + sin (x) + 1)

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the partial fraction of

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

= - \int \frac{1}{u} - \frac{1}{u + 1} d u

Apply the Sum Rule:

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

\int 1 - \frac{2 x}{π} d x

\int \frac{cos ( θ )}{9 sin ^{2} ( θ )} d θ

\int (tan^{3} (x)) (sec^{2} (x)) d x

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