The problem is to find the integration of sqrt(x^2 + 16) .
I have been using x=4tan θ for the substitution and dx=4sec^2 θ .
Please show your work.
With the substitution $\displaystyle \sqrt{x^{2} + 16} = x+z$ and a little computation [...] the integral...
$\displaystyle \displaystyle \int \sqrt{x^{2} + 16}\ dx $ (1)
... becomes the integral of a rational function in z...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
If you draw a right-angled triangle, perpendicular sides...base=x, height=4
then the hypotenuse length is $\displaystyle \sqrt{x^2+4^2}$
$\displaystyle \theta$ is the angle opposite the side of length 4.
$\displaystyle cos\theta=\frac{x}{\sqrt{x^2+16}}\ \Rightarrow\ \sqrt{x^2+16}=\frac{x}{cos\theta}$
$\displaystyle tan\theta=\frac{4}{x}\ \Rightarrow\ x=\frac{4}{tan\theta}$
$\displaystyle \frac{dx}{d\theta}=4\left(-cosec^2\theta\right)$
Therefore,
$\displaystyle \int_{\sqrt{x^2+16}}dx=4\int_{\frac{4}{tan\theta\ cos\theta}}\left(-cosec^2\theta\right)d\theta}$
etc
Hyperbolic substitution may be easier in this case...
Let $\displaystyle x = 4\sinh{t}$ so that $\displaystyle dx = 4\cosh{t}\,dt$.
The integral $\displaystyle \int{\sqrt{x^2 + 16}\,dx}$ becomes
$\displaystyle \int{\sqrt{(4\sinh{t})^2 + 16}\cdot 4\cosh{t}\,dt}$
$\displaystyle = 4\int{\cosh{t}\sqrt{16\sinh^2{t} + 16}\,dt}$
$\displaystyle = 4\int{\cosh{t}\sqrt{16(1 + \sinh^2{t})}\,dt}$
$\displaystyle = 4\int{4\cosh{t}\sqrt{\cosh^2{t}}\,dt}$
$\displaystyle = 16\int{\cosh^2{t}\,dt}$
$\displaystyle = 16\int{\frac{1}{2} + \frac{\cosh{2t}}{2}\,dt}$
$\displaystyle = 16\left(\frac{t}{2} + \frac{\sinh{2t}}{4}\right) + C$
$\displaystyle = 8t + 4\sinh{2t} + C$
$\displaystyle = 8t + 8\sinh{t}\cosh{t} + C$
$\displaystyle = 8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C$.
Now remembering that $\displaystyle x = 4\sinh{t}$ that means $\displaystyle \sinh{t} = \frac{x}{4}$ and $\displaystyle t = \sinh^{-1}\left(\frac{x}{4}\right)$ then
$\displaystyle 8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C = 8\sinh^{-1}\left(\frac{x}{4}\right) + 8\left(\frac{x}{4}\right)\sqrt{1 + \left(\frac{x}{4}\right)^2} +C$
$\displaystyle = 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{1 + \frac{x^2}{16}} +C$
$\displaystyle = 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{\frac{16 + x^2}{16}} + C$
$\displaystyle = 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{2x\sqrt{16 + x^2}}{4} + C$
$\displaystyle = 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{x\sqrt{16 + x^2}}{2} + C$
$\displaystyle = \frac{16\sinh^{-1}\left(\frac{x}{4}\right) + x\sqrt{16 + x^2}}{2} + C$.
if that is the case, then you should have worked your original substitution down to this point ...
$\displaystyle 16 \int \sec^3{t} \, dt$
let's ignore the constant (16) for a bit and focus effort on the integral of secant cubed. using integration by parts ...
$\displaystyle u = \sec{t}$ ... $\displaystyle dv = \sec^2{t} \, dt$
$\displaystyle du = \sec{t} \tan{t} \, dt$ ... $\displaystyle v = \tan{t}$
$\displaystyle \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} \tan^2{t} \, dt
$
$\displaystyle \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} (\sec^2{t} - 1) \, dt$
$\displaystyle \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} - \sec{t} \, dt$
$\displaystyle \int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} \, dt + \int \sec{t} \, dt$
$\displaystyle 2\int \sec^3{t} \, dt = \sec{t} \tan{t} + \int \sec{t} \, dt$
$\displaystyle \int \sec^3{t} \, dt = \frac{1}{2} \sec{t} \tan{t} + \frac{1}{2} \ln|\sec{t} + \tan{t}| + C$
put the constant (16) back ...
$\displaystyle 16\int \sec^3{t} \, dt = 8\sec{t} \tan{t} + 8\ln|\sec{t} + \tan{t}| + C$
I'll leave you the task of back substituting.