# Homework Help: Trig sub

1. Jun 18, 2006

### suspenc3

Hi, IM having trouble with this topic..this is probly all wrong, but any help would be appreciated:

$$\int_{0}^{1} \sqrt{x^2 +1}dx$$

so let

x=tanØ
dx=sec^2Ø

$$\sqrt{x^2+1} = \sqrt{tan^2Ø+ 1} = sec^2Ø$$

when x=0 tanØ = 0 so tanØ = 0
when x=1 tanØ =$$\frac{\pi}{4}[/itex] let u= cosØ du=-sinØ [tex]\int_{0}^{\frac{\pi}{4}} sec^2ØdØ$$=

$$\int_{0}^{\frac{\pi}{4}} \frac{1}{cos^2Ø}dØ$$=

$$-\int_{0}^{\frac{\pi}{4}} \frac{du}{u^2}dØ$$

when Ø = 0 u = 1
when Ø = pi/4, u=root2/2

..i end up with an answer of 1-(root2/2)

Last edited by a moderator: Jun 19, 2006
2. Jun 18, 2006

### StatusX

Could you rewrite some of those equations to make them understandable? The second latex equation, for example, doesn't make sense as is.

Next, you don't seem to be changing dx to du properly, but that may be related to the first thing I said.

Also, I think you want to use hyperbolic functions here, not trigonometric functions. Specifically, the following identities should be useful:

$$\cosh^2 x - \sinh^2 x = 1$$

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

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

Finally, it might be easier to rewrite the integral as:

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

The first term can be done with the right hyperbolic substitution, and the second can be integrated by parts.

Alternatively, you can use the substitution $u=\sqrt{1+x^2}$, but this might be a little trickier.

Last edited: Jun 18, 2006
3. Jun 19, 2006

### Jameson

Mmmk. You had the right idea.

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

$$\tan(\theta)=x$$
$$\sec^2(\theta)d\theta =dx$$

$$\int \sec^2(\theta) \sqrt{\tan^2(\theta)+1}d\theta$$

$$\int \sec^2(\theta) \sqrt{\sec^2(\theta)}d\theta$$

I think you can take it from here.

4. Jun 19, 2006

### HallsofIvy

As Jameson pointed out- you have a number of errors:
$$\sqrt{x^2+1} = \sqrt{tan^2\phi+ 1} = sec^2\phi$$
is, of course, incorrect
$$\sqrt{x^2+1} = \sqrt{tan^2\phi+ 1} = \sqrt{sec^2\phi}= sec\phi$$

$$\int_{0}^{\frac{\pi}{4}} sec^2\phi d\phi$$
$$\int sec^2\phi d\phi= tan \phi$$
$$\phi$$