# Trig substitution integral

## Homework Statement

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

## 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)

LCKurtz
Homework Helper
Gold Member
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.

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.

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

Thanks for the help

LCKurtz