Integrating Sec^3(x) without Absolute Value

[Mr Davis 97]

So I am trying compute $\displaystyle \int \sqrt{1+x^2}dx$. To start, I make the substitution $u=\tan x$. After manipulation, this gives us $\displaystyle \int |\sec u| \sec^2u ~du$. How do I get rid of the absolute value sign, so that I can go about integrating $\sec^3 u$? Is there an argument that shows that $\sec u$ is always positive or always negative?

 

[Svein]

Try substituting x=\sinh(u)...

 

[stevendaryl]

So I am trying compute $\displaystyle \int \sqrt{1+x^2}dx$. To start, I make the substitution $u=\tan x$. After manipulation, this gives us $\displaystyle \int |\sec u| \sec^2u ~du$. How do I get rid of the absolute value sign, so that I can go about integrating $\sec^3 u$? Is there an argument that shows that $\sec u$ is always positive or always negative?

Just a suggestion: Instead of using x = tan(u), you might be better off with x = sinh(u).

As to your original question, sec(u) = \frac{1}{cos(u)}. So sec(u) > 0 whenever cos(u) > 0, which means for |u| < \frac{\pi}{2}. But from your substitution, x = tan(u), there is no reason to consider |u| > \frac{\pi}{2}, because the range -\infty < x < +\infty maps to the range -\frac{\pi}{2} < u < +\frac{\pi}{2}.