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

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

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

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

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

Solution steps

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

Expand

\frac{1 + tan ( x )}{1 - cos ( x )} : \frac{1}{1 - cos ( x )} + \frac{tan ( x )}{1 - cos ( x )}

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

\int \frac{1}{1 - cos ( x )} d x = - cot (\frac{x}{2})

\int \frac{tan ( x )}{1 - cos ( x )} d x = ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1

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

Add a constant to the solution

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

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