What is the integral of (sec(x)-1)^2 ?

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

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integral of (sec(x)-1)^2

\int (sec (x) - 1)^{2} d x

tan (x) - 2 ln ∣ tan (x) + sec (x) ∣ + x + C

Solution steps

\int (sec (x) - 1)^{2} d x

Rewrite using trig identities

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

Expand

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

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

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

\int \frac{2}{cos ( x )} d x = 2 ln ∣ tan (x) + sec (x) ∣

\int 1 d x = x

= tan (x) - 2 ln ∣ tan (x) + sec (x) ∣ + x

Add a constant to the solution

= tan (x) - 2 ln ∣ tan (x) + sec (x) ∣ + x + C

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