Easy? integral

1. Jan 26, 2006

skiboka33

Ok, apparently it's been a while since I've had to deal with an integral like this, seems like it should be easy but I can't find a substitution that will work. The integral is:

$$\int_{0}^{h}\frac{dx}{(x^2+R^2)^\left1/2\right}$$

Where R is a constant.

Any hints (or solutions if you're feeling generous) would be great, thanks

Last edited: Jan 26, 2006
2. Jan 26, 2006

VietDao29

You can either use a subtitution:
x = R tan(u) (you can try it now, if you have any problems, you can post it out here).
If you see some integrals that have:
1. $\sqrt{x ^ 2 + \alpha ^ 2}$, it's common to let $x = \alpha \tan(u)$.
2. $\sqrt{x ^ 2 - \alpha ^ 2}$, it's common to let $x = \alpha \sec(u)$.
3. $\sqrt{\alpha ^ 2 - x ^ 2}$, it's common to let $x = \alpha \sin(u)$.
------------------
Or, you can it in a different way:
Now let: $\sqrt{x ^ 2 + R ^ 2} = t - x$
Differentiate both sides yields:
$$\frac{x \ dx}{\sqrt{x ^ 2 + R ^ 2}} = dt - dx$$.
Rearange it a bit gives:
$$\frac{x + \sqrt{x ^ 2 + R ^ 2}}{\sqrt{x ^ 2 + R ^ 2}} dx = dt$$.
Now, since we have: $\sqrt{x ^ 2 + R ^ 2} = t - x \Leftrightarrow t = \sqrt{x ^ 2 + R ^ 2} + x$, plug it in the expression above gives:
$$\frac{t}{\sqrt{x ^ 2 + R ^ 2}} dx = dt$$
Rearrange it we have:
$$\frac{dx}{\sqrt{x ^ 2 + R ^ 2}} = \frac{dt}{t}$$. Now you can integrate both sides and arrive at the answer you want.
Can you go from here?
Now, let's try the first way to see if they arrive at the same result. (Let x = R tan(u)). Can you go from here?

Last edited: Jan 26, 2006
3. Jan 26, 2006

skiboka33

I actually already attempted to solve using the $$x=Rtan(\theta)$$method before I posted and came up with:

$$\int_{x=0}^{x=h}tan(\theta)sec(\theta)d\theta$$

which i believe gave:

$$\left\frac{1}{cos(\theta)}\right_{x=0}^{x=1}$$

now I know when x=0 tan = 0, but waht about x = h which is a unknown constant? thanks.

EDIT: I just tried the second method, which I had actually never seen before, and it worked, so thanks very much!

Last edited: Jan 26, 2006
4. Jan 26, 2006

VietDao29

I dont' really get what you mean. Can you show me your work?
-----------
Let $$x = R \tan \theta \Rightarrow dx = R \frac{du}{\cos ^ 2 u}$$
Assume that R > 0.
$x = 0 \Rightarrow \theta = 0$
$$x = h \Rightarrow R \tan \theta = h \Rightarrow \tan \theta = \frac{h}{R} \Rightarrow \theta = \arctan \left( \frac{h}{R} \right)$$
So your integral will become:
$$\int \limits_{0} ^ {h} \frac{dx}{\sqrt{x ^ 2 + R ^ 2}} = \int \limits_{0} ^ {\arctan \left( \frac{h}{R} \right)} \frac{R \ d \theta}{R \cos ^ 2 \theta \ \sqrt{1 + \tan ^ 2 \theta}} = \int \limits_{0} ^ {\arctan \left( \frac{h}{R} \right)} \frac{\ d \theta}{\cos ^ 2 \theta \ \sqrt{\frac{1}{\cos ^ 2 \theta}}}$$
$$= \int \limits_{0} ^ {\arctan \left( \frac{h}{R} \right)} \frac{\ d \theta}{\cos \theta}$$.
Is this what you get? Hopefully, you can you from here, right?

Last edited: Jan 26, 2006
5. Jan 28, 2006

Robokapp

Well can't you u-substitute the whole crap under the line?
I'll give it a shot. My Calculus AB should do it...u=x^2+R^2 it is a circle but...meh.

u=x^2+r^2
du/dx=2x so i got rid of your R^2 part.
du = 2x dx
dx = du / 2x

and here we have a small problem...it seems my math made the lower bound fall into a 'hole" because I can't integrate at x=0 however, let's keep going. who knows...bounds do need to change and i still got to get rid of the x.

u=x^2+R^2 so x^2=u-R^2 so x=sqrt(u-R^2)

so dx=du/ sqrt(u-R^2)

$$\int_{r^2}^{h^2+r^2}\frac{du}{(u-R^2)^\left1/2\right}$$

Bounds were found by plugging them nito u equation.

Looks about right...let's continue. integral of all this is a ln so...

Ln(u-R^2) evalauted from R^2 to (h^2+R^2). Or you can change it back and plug the bounds as they are. that means...
Ln(X^2+R^2) from 0 to h
So the answer would be (going with second option because it looks human-like)

Ln(H^2+R^2)-Ln(R^2)

Okay...i have that feeling...that is telling me I wrote a big laod of nonsense...I tried. Don't trust my work...it looked too simple to be right. i'm wrong.

Last edited: Jan 28, 2006
6. Jan 28, 2006

d_leet

Well in your final integral you left out the original part that was supposed to be integrated and just put the dx in.

7. Jan 29, 2006

Robokapp

I did what? Did I forget a "u"? it's 3:30 am...so it's likely. can you ponit out plz?

8. Jan 29, 2006

d_leet

I'm tired to soo I'm not going to post the actual integral but you found dx in terms of u and du and that is all you put into the final integral you forgot the original integrand in terms of u.

9. Jan 29, 2006

arildno

The simplest way of cracking this integral is to use the hyperbolic substitution $x=Rsinh(u)$

That's just my opinion, though.

Similar Discussions: Easy? integral
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9830921292304993, "perplexity": 1582.908679305428}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608956.34/warc/CC-MAIN-20170527152350-20170527172350-00012.warc.gz"}