trig sub

suspenc3

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

[tex]\int_{0}^{1} \sqrt{x^2 +1}dx[/tex]

so let

x=tanØ
dx=sec^2Ø

[tex]\sqrt{x^2+1} = \sqrt{tan^2Ø+ 1} = sec^2Ø[/tex]

when x=0 tanØ = 0 so tanØ = 0
when x=1 tanØ =[tex]\frac{\pi}{4}[/itex]

let u= cosØ
du=-sinØ

[tex]\int_{0}^{\frac{\pi}{4}} sec^2ØdØ[/tex]=

[tex]\int_{0}^{\frac{\pi}{4}} \frac{1}{cos^2Ø}dØ[/tex]=

[tex]-\int_{0}^{\frac{\pi}{4}} \frac{du}{u^2}dØ[/tex]

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

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

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:

[tex] \cosh^2 x - \sinh^2 x = 1 [/tex]

[tex]\frac{d}{dx} (\sinh x ) = \cosh x [/tex]

[tex]\frac{d}{dx} (\cosh x ) = \sinh x [/tex]

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

[tex] \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[/tex]

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 [itex]u=\sqrt{1+x^2}[/itex], but this might be a little trickier.

Jameson

Mmmk. You had the right idea.

[tex]\int \sqrt{x^2+1}dx[/tex]

[tex]\tan(\theta)=x[/tex]
[tex]\sec^2(\theta)d\theta =dx[/tex]

[tex]\int \sec^2(\theta) \sqrt{\tan^2(\theta)+1}d\theta[/tex]

[tex]\int \sec^2(\theta) \sqrt{\sec^2(\theta)}d\theta[/tex]

I think you can take it from here.

HallsofIvy

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

Actually, if it had been correct and you had
[tex]\int_{0}^{\frac{\pi}{4}} sec^2\phi d\phi[/tex]
it would have been easy!
[tex]\int sec^2\phi d\phi= tan \phi[/tex]

The substitution you made was incorrect because you forgot if u= cos Ø
then du= -sin Ø

By the way, "alt" codes do not work inside LaTex you need \phi to get
[tex]\phi[/tex]