# Trig substitution: integrate sqrt(16+x^2) over x

1. Jan 31, 2012

### skyturnred

1. The problem statement, all variables and given/known data

$\int\frac{\sqrt{16+x^{2}}}{x}$

2. Relevant equations

3. The attempt at a solution

set x=4tant
dx=4sec$^{2}$t dt

so after plugging in and using a quick trig identity I get:

$\int\frac{16(sec^{2}t)*4sec^{2}t dt}{4tant}$

Then after a quick cleanup:

16*$\int tan^{-1}t*sec^{4}t *dt$

I think I made a mistake, but if not, what next?

2. Jan 31, 2012

### LCKurtz

Best not to use $\tan^{-1}$ for $1/\tan$ since that may be confused with arctan. But write $\sec^4(x)= \sec^2 x(1 + \tan^2 x)$ and try a u - substitution $u=\tan x$.

3. Jan 31, 2012

### jimbobian

I think you've made a mistake, where has the square root gone?

4. Jan 31, 2012

### skyturnred

I definitely made a mistake. I forgot the square root sign, which changes everything. Thanks!