# How do you integrate sec^3(x)

1. Jan 19, 2009

### The_ArtofScience

This is not homework, but I'm just wondering, how do you integrate this deceptive looking integrand to get what Wolfram has?

I don't get why the answer has an inverse hyperbolic function. Please teach me!

2. Jan 19, 2009

### rock.freak667

$$\int sec^3xdx = \int secx (sec^2x dx)$$

Integration by parts and then use the identity $sec^2x=tan^2x+1$

3. Jan 19, 2009

### The_ArtofScience

That method leads to (1/2)sec(x)tan(x) + (1/2)ln(sec(x) + tan(x)) + C. I am interested in getting an inverse hyperbolic function as displayed on Wolfram.

I do not know how inverse hyperbolic functions are related to integrals. The only success I've had was integrating sec(x) into 2tanh^-1(tan(x/2))

4. Jan 19, 2009

### NoMoreExams

Maybe you should post what wolfram got?

5. Jan 19, 2009

### The_ArtofScience

6. Jan 20, 2009

### Mute

Inverse hyperbolic functions can be written in terms of logarithms. In particular,

$$\operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$

So playing around with your logarithm you can probably get the artanh function they give out. (You may need to add a constant to your result to get to theirs).