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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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} ( t )} d t - \int \frac{2}{cos ( t )} d t + \int 1 d t

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

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

\int 1 d t = t

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

Add a constant to the solution

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

a re a y = x^{3}, [0, 4]

d er i v a t i v e \frac{6}{x}

\int (- 9 x^{2} + 9 x + 2) d x

d er i v a t i v e y = (x^{3} + 8 x^{2})^{- 1}

d er i v a t i v e f (x) = \frac{2 x + 5}{x - 1}

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