# I Integrating sqrt(1+x^2)

1. Jan 23, 2017

### 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?

2. Jan 23, 2017

### Svein

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

3. Jan 23, 2017

### stevendaryl

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

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