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

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

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

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

tan (x) - x + C

Solution steps

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

\int sec^{2} (x) d x = tan (x)

\int 1 d x = x

= tan (x) - x

Add a constant to the solution

= tan (x) - x + C

x \to - 2 lim (2 x + 9)

\int x sin (- 1 x) d x

\int e^{- y} d y

F (x) = sin (tan (3 x))

\frac{d}{d x} (ln (3 x + 1))

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