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

\int_{0}^{\frac{π}{2}} cos^{3} (x) d x

\frac{d}{d x} (sin (x - 3))

x \to 2 lim (\frac{3 x}{x ^{2} + 2})

\frac{\partial}{\partial y} (2 x^{4} y + 1)

y = 4 x - x^{2}, y = 3 x

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