integrate 1/cos(x)^2

1) Let's find the domain of

\frac{1}{{\cos (x)}^{2}}

⇥ For the expression

\frac{1}{{\cos (x)}^{2}}

to be defined, we must have:

{\cos (x)}^{2} \neq 0

⇥From

{\cos (x)}^{2} \neq 0

, we deduce:

\cos (x) \neq 0

⇥From cos(θ) ≠

0

, we deduce: θ ≠

\frac{π}{2} + k π

, k ∈ ℤ

⇥Solution obtained:

x \neq \frac{π}{2} + k π

⇥Domain of definition:

{x \neq \frac{π}{2} + k π, k \in ℤ}

2) Let's compute the integral:

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

ⓘ The derivative of

\tan (x)

\frac{1}{{\cos (x)}^{2}}

, so

\tan (x)

is an antiderivative of

\frac{1}{{\cos (x)}^{2}}

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

\tan (x)

\tan (x)

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\tan (x) + C

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