# Trigonometric Integral Confusion

1. Jul 12, 2010

### Char. Limit

1. The problem statement, all variables and given/known data

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 cancelled 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.

2. Relevant equations

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

3. 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))?

2. Jul 12, 2010

### Bohrok

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

3. Jul 12, 2010

### vela

Staff Emeritus
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.

4. Jul 12, 2010

### Char. Limit

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)...

5. Jul 12, 2010

### Bohrok

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

6. Jul 12, 2010

### vela

Staff Emeritus
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.

7. Jul 12, 2010

### 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.

8. Jul 13, 2010

### 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.

9. Jul 13, 2010

### 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?

10. Jul 13, 2010

### 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

11. Jul 13, 2010

### Char. Limit

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

12. Jul 13, 2010

### 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

13. Jul 13, 2010

### Char. Limit

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