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

The integral of (sec(θ)tan(θ))/(1-sec(θ)) is ln|cos(θ)|-ln|cos(θ)-1|+C

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integral of (sec(θ)tan(θ))/(1-sec(θ))

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

ln ∣ cos (θ) ∣ - ln ∣ cos (θ) - 1 ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{tan ( θ )}{cos ( θ ) - 1} d θ

Apply u-substitution

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

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

Take the constant out:

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

= - \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 = cos (θ)

= - (- ln ∣ cos (θ) ∣ + ln ∣ cos (θ) - 1 ∣)

Simplify

= ln ∣ cos (θ) ∣ - ln ∣ cos (θ) - 1 ∣

Add a constant to the solution

= ln ∣ cos (θ) ∣ - ln ∣ cos (θ) - 1 ∣ + C

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