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

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

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

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

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

Solution steps

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

Add a constant to the solution

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

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