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

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

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

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

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

Solution steps

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

Add a constant to the solution

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

x \to - \infty lim (a + \frac{1}{x})

(7 x^{2} + 6 x) (7 x - 8)

f (x) = x^{2} + 2 x

\frac{d}{d x} (\frac{x ^{2} + 6 x + 9}{x - 3})

f (x) = (2 x - 5) (3 x + 1)

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