# Indefinite integral of arcsec(x)

1. May 14, 2015

### PWiz

Just for fun, I tried this rather trivial problem, but I think I went wrong somewhere:
$$\int arcsec(x) \ dx$$
Let $arcsec(x)=y$ . Then $x=sec \ y$, or $y=arccos(\frac 1{x})$
So the problem becomes $$\int arccos(\frac 1 {x}) \ dx$$
Let $\frac 1 {x} = cos \ u$ , so that $dx = secu \ tanu \ du$.
So the questions simplifies to $\int u \ secu \ tanu \ du$. If I integrate by parts, then this equals to
$$u \int secu \ tanu \ du - \int (\int secu \ tanu \ du) du$$
Since $\int secu \ tanu \ du = secu$, I get $u \ secu - \int secu \ du$
$\int secu \ du = ln(\sqrt{\frac{1+sinu}{1-sinu}})$
Substituting the Xs, I get $$x \ arcsec(x) - ln(\sqrt{\frac{1+\sqrt{1-\frac{1}{x^2}}}{1-\sqrt{1-\frac{1}{x^2}}}}) + c= x \ arcsec(x) - ln(\sqrt{\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}}) + c$$
where $c$ is an arbitrary constant. I checked my answer with Wolfram Alpha here:
http://integrals.wolfram.com/index.jsp?expr=arcsec(x)&random=false
As you can clearly see, the answer is different from mine. In which step did I goof?

2. May 14, 2015

### Raghav Gupta

Your all steps are correct, you have not goofed.