I can't see how to integrate (1 + t^2)/(1/2)

[TheOli]

I am revising for my exams and am currently looking at the derivatives of hyperbolic functions in preperation for my calculus exam. In the practice exam questions im doing on this subject the following came up:

Integrate the following:

(1+t^2)^(1/2)

I know the answer to be:

1/2 (x Sqrt[1 + x^2] + ArcSinh[x])

I cant see how the answer is found or what this has to do with hyperbolic functions, other than the arcsinh in the answer of course.

 

[BigGlenntheHeavy]

Re: I can't see how to integrate this

Let t = tan(x), then dt = [sec(x)]^2dx, Can you take it from here?

 

[TheOli]

Re: I can't see how to integrate this

Thanks, thats a great first step. I can never get the substitutions.

But now im stuck with integrating (Sec x)^3 which is where i ended up after the substitution. Have i gone wrong, I just cant see where im going to get the hypobolic function from.

 

[BigGlenntheHeavy]

Re: I can't see how to integrate this

Int[sec(x)]^3 dx = [sec(x)tan(x)]/2 + (1/2)ln|sec(x) + tan(x)| + C

Now note: t = tan(x ), x = arctan(t).

 

[DrMike]

Re: I can't see how to integrate this

Let t = tan(x), then dt = [sec(x)]^2dx, Can you take it from here?

Would it not be easier to take t = sinh(x), so dt=cosh(x)dx ?

 

[BigGlenntheHeavy]

Re: I can't see how to integrate this

Dr. Mike, as always, there is more than one way to skin a cat. I usually avoid hyperbolic functions whenever possible.