## trig sub

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)
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Recognitions: Homework Help 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.
 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.

Recognitions:
Gold Member
Staff Emeritus

## trig sub

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

$$\int_{0}^{\frac{\pi}{4}} sec^2\phi d\phi$$
it would have been easy!
$$\int sec^2\phi d\phi= tan \phi$$

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
$$\phi$$