# Trig substitution integral

1. Feb 4, 2013

### iRaid

1. The problem statement, all variables and given/known data
$$\int \frac{dx}{\sqrt{x^{2}+16}}$$

2. Relevant equations

3. The attempt at a solution
$$x=4tan\theta$$ $$dx=4sec^{2}\theta d\theta$$
Therefore:
$$\int \frac{4sec^{2}\theta d\theta}{\sqrt{16tan^{2}\theta +16}} = \int \frac{sec^{2}\theta d\theta}{\sqrt{tan^{2}\theta+1}}$$
$$\int \frac{sec^{2}\theta d\theta}{sec\theta} = \int sec\theta d\theta$$
$$=ln|sec\theta+tan\theta| = ln|sec\frac{\sqrt{x^{2}+16}}{4}+tan\frac{x}{4}|$$

Wolfram is getting a hyperbolic sine function so idk what is wrong (we've never talked about hyperbolic functions in class, so I don't think that's the answer)

2. Feb 4, 2013

### LCKurtz

You are OK at $\ln|\sec\theta+\tan\theta|$. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

Don't worry about the hyperbolic functions. Integrals like this can be worked either way.

3. Feb 4, 2013

### iRaid

Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

Thanks for the help

4. Feb 4, 2013

### LCKurtz

They can be easier, but in this case I don't think it makes much difference. Try $x = 4\sinh t$ on it and see what you think. Remember the basic hyperbolic identity is $\cosh^2t - \sinh^2 t = 1$.