- #1

PWiz

- 695

- 116

$$\int arcsec(x) \ dx$$

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?