# Integral of Square root

• Sep 25th 2008, 11:54 AM
diffeqhelp
Integral of Square root
I have two problems that I'm having trouble with. Both of them involve square roots in the denominator. I'm not sure how to deal with that. Also, I'm new to the forum so I'm not sure how to make all the math symbols appear. Please bear with me! (Tongueout)

1) integrate (1/(x^2-r^2)^(1/2))dx assuming r is some constant

2) integrate ((x+2)/((x^2+4x+5)^(1/2)))dx from 0 to infinity

I'm really stuck, so I'd appreciate any help!!
• Sep 25th 2008, 01:13 PM
Chris L T521
Quote:

Originally Posted by diffeqhelp
I have two problems that I'm having trouble with. Both of them involve square roots in the denominator. I'm not sure how to deal with that. Also, I'm new to the forum so I'm not sure how to make all the math symbols appear. Please bear with me! (Tongueout)

1) integrate (1/(x^2-r^2)^(1/2))dx assuming r is some constant

2) integrate ((x+2)/((x^2+4x+5)^(1/2)))dx from 0 to infinity

I'm really stuck, so I'd appreciate any help!!

For $\int\frac{\,dx}{\sqrt{x^2-r^2}}$, use a trig substitution: $x=r\sec\vartheta$

For $\int\frac{x+2}{\sqrt{x^2+4x+5}}\,dx$, make the u substitution $u=x^2+4x+5$

Try these and see how far you get. If you get stuck, ask us :D

--Chris
• Sep 25th 2008, 08:22 PM
diffeqhelp
Thanks!!
The u-substitution for #2 worked very well. I obtained sqrt(x^2+4x+5)+c, as an indefinite solution. From 0 to infinity, I got, infinity-sqrt(5). (which I figure is still infinity, but I guess I'll write in the sqrt(5) for completeness)

I'm still confused about #1 though. This is what I have so far. Using the trig substitution, x=r*sec(theta); dx=rsec(theta)tan(theta)d(theta). However, I'm unsure what to substitute 1/sqrt(x^2-r^2) with. I thought maybe (sin(theta))/x. But I don't know how that helps? (Wondering)
• Sep 25th 2008, 08:32 PM
o_O
$x = r\sec \theta \ \Rightarrow \ dx = r\sec \theta \tan \theta \ d \theta$

Making our subs: $\int \frac{dx}{\sqrt{x^2 - r^2}} \ = \ \int \frac{r \sec \theta \tan \theta \ d \theta}{\sqrt{r^2\sec^2 \theta - r^2}}$

The reason why we used that sub is so that we can use the identity: $\tan^2 \theta = \sec^2 \theta - 1$

See if you can work with that by first factoring out the $r^2$ ;)
• Sep 25th 2008, 10:37 PM
diffeqhelp
That was a ton of help! Thank you!! (Clapping) I still ran into a rough spot...and I'll show you where it got messy, so I'm not completely sure that my answer is 100% right.

so, starting from where o_O left off, the integral of r*sec(theta)*tan(theta)d(theta)/sqrt((r^2)(sec^2(theta)-1) = the integral of (r*sec(theta)*tan(theta)d(theta))/sqrt((r^2)-tan^2(theta) = the integral of (r*sec(theta)*tan(theta)d(theta))/(r*tan(theta)) = the integral of sec(theta)d(theta)
at which point I googled some help, which gave me a very detailed solution, that I probably wouldn't have thought of on my own.

getting back to my eventual answer, I end up with ln|sec(theta) + tan(theta)| + C. But then I need to convert back to x and r.

I find that theta = arcsec(x/r), through the first trig substitution that Chris L T521 gave me.

substituting back in, gives me ln|(x/r) + tan(arcsec(x/r)|+C
I'm not sure that this is the simpliest form?

Whew that was long..I really need to figure out how to work these math symbols so it doesn't look so messy....Thanks again!
• Sep 26th 2008, 02:20 PM
Chris L T521
Quote:

Originally Posted by diffeqhelp
That was a ton of help! Thank you!! (Clapping) I still ran into a rough spot...and I'll show you where it got messy, so I'm not completely sure that my answer is 100% right.

so, starting from where o_O left off, the integral of r*sec(theta)*tan(theta)d(theta)/sqrt((r^2)(sec^2(theta)-1) = the integral of (r*sec(theta)*tan(theta)d(theta))/sqrt((r^2)-tan^2(theta) = the integral of (r*sec(theta)*tan(theta)d(theta))/(r*tan(theta)) = the integral of sec(theta)d(theta)
at which point I googled some help, which gave me a very detailed solution, that I probably wouldn't have thought of on my own.

getting back to my eventual answer, I end up with ln|sec(theta) + tan(theta)| + C. But then I need to convert back to x and r.

I find that theta = arcsec(x/r), through the first trig substitution that Chris L T521 gave me.

substituting back in, gives me ln|(x/r) + tan(arcsec(x/r)|+C
I'm not sure that this is the simpliest form?

Whew that was long..I really need to figure out how to work these math symbols so it doesn't look so messy....Thanks again!

Note that $x=r\sec\vartheta$

This means that $\sec\vartheta=\frac{x}{r}$

Now when we're dealing with right angle trigonometry, this means that the hypotenuse will have a length of $x$ and the side adjacent to the angle will have a length of $r$. Thus, the side opposite the angle will have a length of $\sqrt{x^2-r^2}$

So the tangent of this angle will be opposite over adjacent, or $\tan\vartheta=\frac{\sqrt{x^2-r^2}}{r}$

So you could write your answer as $\ln\left|\frac{x}{r}+\frac{\sqrt{x^2+r^2}}{r}\righ t|+C=-\ln\left|r\left(x+\sqrt{x^2-r^2}\right)\right|+C$

Does this make sense?

--Chris
• Sep 26th 2008, 04:48 PM
Krizalid
Actually, $\ln r$ is a constant, hence combine constants.
• Sep 26th 2008, 05:05 PM
Chris L T521
Quote:

Originally Posted by Krizalid
Actually, $\ln r$ is a constant, hence combine constants.

Gah! I didn't see that...

So the solution would then take on the form $-\ln\left|x+\sqrt{x^2-r^2}\right|+C$

--Chris
• Sep 30th 2008, 08:09 PM
diffeqhelp
That's really cool. How would I know though that $\sec\vartheta=\frac{x}{r}$ vs. $\csc\vartheta=\frac{x}{r}$? Besides that it makes the solution work?

Quote:

Originally Posted by Chris L T521
Note that $x=r\sec\vartheta$

This means that $\sec\vartheta=\frac{x}{r}$

Now when we're dealing with right angle trigonometry, this means that the hypotenuse will have a length of $x$ and the side adjacent to the angle will have a length of $r$. Thus, the side opposite the angle will have a length of $\sqrt{x^2-r^2}$

So the tangent of this angle will be opposite over adjacent, or $\tan\vartheta=\frac{\sqrt{x^2-r^2}}{r}$

So you could write your answer as $\ln\left|\frac{x}{r}+\frac{\sqrt{x^2+r^2}}{r}\righ t|+C=-\ln\left|r\left(x+\sqrt{x^2-r^2}\right)\right|+C$

Does this make sense?

--Chris