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:


 * $\displaystyle \int \operatorname{arcsec}x \ \mathrm dx = x\operatorname{arcsec}x - \int \frac x \ \mathrm dx$

Suppose $x > 1$.

Then:

Suppose $x < -1$.

Then: