Trigonometric Integral Confusion

[Char. Limit]

Homework Statement

So, for absolutely no good reason I decided to practice trigonometric substitution. I started with $\int \left(1-x^2\right)^{-1/2} dx$, and that was easy, everything canceled out nicely. Then I tried $\int \left(1+x^2\right)^{-1/2} dx$, and although the integral was fine, everything did not cancel out nicely afterward. So I need help on one minor issue.

Homework Equations

$$sec(u) = \frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)}$$

The Attempt at a Solution

$$\int \frac{dx}{\sqrt{1+x^2}}$$

$$\int \frac{sec^2(u) du}{\sqrt{1+tan^2(u)}}$$

$$\int sec(u) du$$

$$\int \frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)} du$$

$$\int \frac{dv}{v}$$

$$ln(v)$$

$$ln(tan(u) + sec(u))$$

$$ln(x + sec(tan^{-1}(x)))$$



Now, is there any way to simplify sec(arctan(x))?

 

[Bohrok]

Let θ = arctan(x), then tan(θ) = x = x/1. Draw a right triangle, label the sides, and then find sec(θ).

 

[vela]

Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1.

 

[Char. Limit]

Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1.

Hmm... I did that, but I don't think my answer is right... the O=x, A=1, H=sqrt(1+x^2) idea gives me sec(arctan(x))=sqrt(1+x^2), and that just doesn't sound right... but it probably is.

So...

$$ln\left(x+\sqrt{1+x^2}\right) +C$$

Is the answer? For some reason I expected a trig function to pop up like it did with 1/sqrt(1-x^2)...

 

[Bohrok]

That's the right answer, you can always check on Wolfram|Alpha too.

 

[vela]

Or you could differentiate your answer and see if you get the integrand back.

If you think in terms of the hyperbolic trig functions, your answer is arcsinh(x), so a trig function does actually pop up.

 

[Char. Limit]

Oh I see. Thanks for the info on the arsinh(x) function... I just looked it up on Wikipedia, and it helped me understand that there was a trig function hidden in that logarithmic function.

 

[hunt_mat]

There are two ways to solve this integral. There is the use of $$x=\tan\theta$$ and by the use of the identity $$1+\tan^{2}theta =\sec^{2}\theta$$ and the use of $$x=\sinh u$$ and use $$\cosh^{2}u=1+\sinh^{2}u$$. I personally think it is easier with the hyperbolic function.

 

[Char. Limit]

Yes, but I had never heard of the hyperbolic functions before today, so I don't quite know the derivative of sinh and cosh. Is it the same as the derivative of sin and cos? Namely...

$$\frac{d sinh(x)}{dx} = cosh(x)$$

$$\frac{d cosh(x)}{dx} = - sinh(x)$$

Are those correct?

 

[hunt_mat]

No. The definitions are as follows:
$$\sinh x=\frac{e^{x}-e^{-x}}{2}\quad\cosh x=\frac{e^{x}+e^{-x}}{2}$$
From these definitions, you can calculate the derivatives and also prove the basic identity:
$$\cosh^{2}x-\sinh^{2}x=1$$
From these three definitions, you can pretty much prove everything you need.

Mat

 

[Char. Limit]

Oh, I see. So the derivative is a two-point cycle, not a four-point cycle. Thank you.

 

[hunt_mat]

There's not much to learn about hyperbolic functions apart from their definitions, their graphs and their derivatives. I am not sure of their geometric interpretation, there must be one, hyperbolic geometry perhaps?

mat

 

[Char. Limit]

I just looked it up, they are supposed to give measures of a hyperbola, I believe. Somehow.