What is the integral of sec^2(θ)-1 ?

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

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integral of sec^2(θ)-1

\int sec^{2} (θ) - 1 d θ

tan (θ) - θ + C

Solution steps

\int sec^{2} (θ) - 1 d θ

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} (θ) d θ - \int 1 d θ

\int sec^{2} (θ) d θ = tan (θ)

\int 1 d θ = θ

= tan (θ) - θ

Add a constant to the solution

= tan (θ) - θ + C

\int e^{x} - \frac{2}{x} d x

\int_{- 3}^{0} (x^{2} - 4 x + 7) d x

\int (x + 3) 2 - x d x

x \to 4 - lim (5 (x) - 7)

d er i v a t i v e arccos (1 - x^{2})