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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= - 2 \cdot \int \frac{1}{u - 1} - \frac{u}{u ^{2} + 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

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

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

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

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

Substitute back

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

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

Simplify

- 2 (ln tan (\frac{x}{2}) - 1 - \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1) : - 2 ln - 1 + tan (\frac{x}{2}) + ln (sec^{2} (\frac{x}{2}))

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

Add a constant to the solution

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

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