How can logarithms be used to simplify inverse hyperbolic functions?

[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!

 

[rock.freak667]

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

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

 

[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))

 

[NoMoreExams]

Maybe you should post what wolfram got?

 

[The_ArtofScience]

Oh, sorry about that, here it is:

http://integrals.wolfram.com/index.jsp?expr=sec^3(x)&random=false

tanh^-1(tan(x/2)) + 1/(4 - 4sin(x)) - 1/(4(cos(x/2) + sin(x/2))^2)

 

[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).