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 $-\frac{\pi}{2} < u < +\frac{\pi}{2}$.