Math Help - I need help integrating using trig substitutions

1. I need help integrating using trig substitutions

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 .

2. With the substitution $\sqrt{x^{2} + 16} = x+z$ and a little computation [...] the integral...

$\displaystyle \int \sqrt{x^{2} + 16}\ dx$ (1)

... becomes the integral of a rational function in z...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by mlhuff12
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 .

If you draw a right-angled triangle, perpendicular sides...base=x, height=4

then the hypotenuse length is $\sqrt{x^2+4^2}$

$\theta$ is the angle opposite the side of length 4.

$cos\theta=\frac{x}{\sqrt{x^2+16}}\ \Rightarrow\ \sqrt{x^2+16}=\frac{x}{cos\theta}$

$tan\theta=\frac{4}{x}\ \Rightarrow\ x=\frac{4}{tan\theta}$

$\frac{dx}{d\theta}=4\left(-cosec^2\theta\right)$

Therefore,

$\int_{\sqrt{x^2+16}}dx=4\int_{\frac{4}{tan\theta\ cos\theta}}\left(-cosec^2\theta\right)d\theta}$

etc

4. Originally Posted by mlhuff12
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 .

Hyperbolic substitution may be easier in this case...

Let $x = 4\sinh{t}$ so that $dx = 4\cosh{t}\,dt$.

The integral $\int{\sqrt{x^2 + 16}\,dx}$ becomes

$\int{\sqrt{(4\sinh{t})^2 + 16}\cdot 4\cosh{t}\,dt}$

$= 4\int{\cosh{t}\sqrt{16\sinh^2{t} + 16}\,dt}$

$= 4\int{\cosh{t}\sqrt{16(1 + \sinh^2{t})}\,dt}$

$= 4\int{4\cosh{t}\sqrt{\cosh^2{t}}\,dt}$

$= 16\int{\cosh^2{t}\,dt}$

$= 16\int{\frac{1}{2} + \frac{\cosh{2t}}{2}\,dt}$

$= 16\left(\frac{t}{2} + \frac{\sinh{2t}}{4}\right) + C$

$= 8t + 4\sinh{2t} + C$

$= 8t + 8\sinh{t}\cosh{t} + C$

$= 8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C$.

Now remembering that $x = 4\sinh{t}$ that means $\sinh{t} = \frac{x}{4}$ and $t = \sinh^{-1}\left(\frac{x}{4}\right)$ then

$8t + 8\sinh{t}\sqrt{1 + \sinh^2{t}} + C = 8\sinh^{-1}\left(\frac{x}{4}\right) + 8\left(\frac{x}{4}\right)\sqrt{1 + \left(\frac{x}{4}\right)^2} +C$

$= 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{1 + \frac{x^2}{16}} +C$

$= 8\sinh^{-1}\left(\frac{x}{4}\right) + 2x\sqrt{\frac{16 + x^2}{16}} + C$

$= 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{2x\sqrt{16 + x^2}}{4} + C$

$= 8\sinh^{-1}\left(\frac{x}{4}\right) + \frac{x\sqrt{16 + x^2}}{2} + C$

$= \frac{16\sinh^{-1}\left(\frac{x}{4}\right) + x\sqrt{16 + x^2}}{2} + C$.

5. I haven't learned anything about hyperbolic substitution yet. In my notes it says in the case of sqrt(x²+a²) to use atan θ as the substitution.

6. You will get an equivalent answer (because you can convert $\sinh^{-1}\left(\frac{x}{4}\right)$ to its logarithmic equivalent, but using trigonometric substitution in this case gets messy.

7. Originally Posted by mlhuff12
I haven't learned anything about hyperbolic substitution yet. In my notes it says in the case of sqrt(x²+a²) to use atan θ as the substitution.
if that is the case, then you should have worked your original substitution down to this point ...

$16 \int \sec^3{t} \, dt$

let's ignore the constant (16) for a bit and focus effort on the integral of secant cubed. using integration by parts ...

$u = \sec{t}$ ... $dv = \sec^2{t} \, dt$

$du = \sec{t} \tan{t} \, dt$ ... $v = \tan{t}$

$\int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} \tan^2{t} \, dt
$

$\int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec{t} (\sec^2{t} - 1) \, dt$

$\int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} - \sec{t} \, dt$

$\int \sec^3{t} \, dt = \sec{t} \tan{t} - \int \sec^3{t} \, dt + \int \sec{t} \, dt$

$2\int \sec^3{t} \, dt = \sec{t} \tan{t} + \int \sec{t} \, dt$

$\int \sec^3{t} \, dt = \frac{1}{2} \sec{t} \tan{t} + \frac{1}{2} \ln|\sec{t} + \tan{t}| + C$

put the constant (16) back ...

$16\int \sec^3{t} \, dt = 8\sec{t} \tan{t} + 8\ln|\sec{t} + \tan{t}| + C$

I'll leave you the task of back substituting.

8. Thanx. So apparently I already had the right answer. However completely different than the back of my book says though.

9. the "right" answer is in terms of x ... I told you that my work was incomplete.