Integral of Arcsecant Function

Theorem

$\displaystyle \int \operatorname{arcsec}x \ \mathrm dx = x\operatorname{arcsec}x - \ln \left \vert{x + \sqrt {x^2 - 1} } \right \vert + C$

for $x^2 > 1$.

Proof

$\displaystyle \int \operatorname{arcsec}x \ \mathrm dx = \int 1 \cdot \operatorname{arcsec}x \ \mathrm dx$

From Integration by Parts:

$\displaystyle \int f'(x)g(x) \ \mathrm dx = f(x)g(x) - \int f(x)g'(x) \ \mathrm dx$

where:

• $f'(x) = 1$

• $f(x) = x$

• $g(x) = \operatorname{arcsec} x$

• $g'(x) = \dfrac 1 {|x|\sqrt{x^2 -1}}$

we then have:

$\displaystyle \int \operatorname{arcsec}x \ \mathrm dx = x\operatorname{arcsec}x - \int \frac x {{|x|\sqrt{x^2 -1}}} \ \mathrm dx$

Suppose $x > 1$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \frac x { {\left \vert{x}\right \vert} \sqrt{x^2 -1} } \ \mathrm dx\)	\(=\)	\(\displaystyle \int \frac 1 { {\sqrt{x^2 - 1} } } \ \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Absolute Value
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ln \left({x + \sqrt {x^2 - 1} } \right) + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integral of One Over Square Root of Binomial
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ln \left \vert{x + \sqrt {x^2 - 1} } \right \vert + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Absolute Value

Suppose $x < -1$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \frac x { {\left \vert{x}\right \vert} \sqrt{x^2 -1} } \ \mathrm dx\)	\(=\)	\(\displaystyle -\int \frac 1 { {\sqrt{x^2 - 1} } } \ \mathrm dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Absolute Value
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -1 \cdot -\ln \left({-x - \sqrt {x^2 - 1} } \right) + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integral of One Over Square Root of Binomial
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \ln \left \vert{x + \sqrt {x^2 - 1} } \right \vert + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Absolute Value

$\blacksquare$