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)-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}) - 1 + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v} d v

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

\int_{0}^{\frac{7}{3}} 9 + 3 x d x

\frac{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

\int \frac{( 5 t - t )}{3 t ^{2} - 9} d t

\frac{\partial}{\partial x} (x^{4} - 3 x^{2} y^{2} + y^{4})

x \to - 2 lim (4^{2} - 3)

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)-1|+C