What is the integral of (sec(x))/(2tan(x)+sec(x)-1) ?

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

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integral of (sec(x))/(2tan(x)+sec(x)-1)

\int \frac{sec ( x )}{2 tan ( x ) + sec ( x ) - 1} d x

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

Solution steps

\int \frac{sec ( x )}{2 tan ( x ) + sec ( x ) - 1} d x

Apply u-substitution

: \int \frac{1}{u ( u + 2 )} d u

= \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Substitute back

u = tan (\frac{x}{2})

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

Add a constant to the solution

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

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