# Improper Integrals!!!!!!

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• Jun 21st 2008, 04:53 PM
elocin
Improper Integrals!!!!!!
I need help integrating from 0 to 1, the integral of (41) all over (square root of 1-x^2). I think that this integral converges, but I'm not sure if I'm evaluating the integral right, I started by putting the 41 in front of the integral and maybe using trig substitution?! Ah please help me, I am out of it, with math!
• Jun 21st 2008, 05:03 PM
Krizalid
$\int{\frac{dx}{\sqrt{1-x^{2}}}}=\arcsin x+k,$ which is a standard integral.

Now if you wanna prove such result, make the substitution $x=\sin\varphi,\,-\frac\pi2\le\varphi\le\frac\pi2.$
• Jun 21st 2008, 05:31 PM
elocin
Okay! I used x=sintheta and so I got (1) over the square root of (1-(sintheta squared) squared. That looks kind of weird, but then I got rid of the square root and got (1) over (1-sin(theta)^2). That is the same as (1) over cos(theta)^2, which is the same thing as secant squared, and then integrating that gets you tangent I think. So, I'm that far. I think I have to use the triangle, to change theta back to x. I think that is opposite over adjacent, which is (x) over (the square root of 1-x^2). I'm not sure where to go from there.
• Jun 21st 2008, 05:35 PM
topsquark
Quote:

Originally Posted by elocin
Okay! I used x=sintheta and so I got (1) over the square root of (1-(sintheta squared) squared.

There's your mistake. What you have written is
$\int \frac{1}{ \left ( \sqrt{1 - sin^2(\theta)} \right ) ^2}~dx$

You have an extra "squared" in there. It should be
$\int \frac{1}{\sqrt{1 - sin^2(\theta)}}~dx$

Also,
$x = sin(\theta) \implies dx = cos(\theta)~d \theta$
You need to replace the differential element as well.

-Dan
• Jun 21st 2008, 05:54 PM
elocin
Hmm, okay So I have the integral of (1) over the square root of 1-sintheta squared and cos(theta) dtheta. I simplified that to be (1) over the square root cosine squared, cosine theta d theta. So I simplified that to be (1) over cosine theta, cosine theta d theta. I canceled out a cosine theta, and therefore got 1 dtheta. Can that be right? Integrating that, I would get just theta?
• Jun 21st 2008, 05:58 PM
Chris L T521
Quote:

Originally Posted by elocin
Hmm, okay So I have the integral of (1) over the square root of 1-sintheta squared and cos(theta) dtheta. I simplified that to be (1) over the square root cosine squared, cosine theta d theta. So I simplified that to be (1) over cosine theta, cosine theta d theta. I canceled out a cosine theta, and therefore got 1 dtheta. Can that be right? Integrating that, I would get just theta?

Yes. You should get $\theta+K$. But what is theta? Recall that you made the substitution $x=\sin(\theta)$...

--Chris
• Jun 21st 2008, 06:05 PM
elocin
I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!
• Jun 21st 2008, 06:09 PM
topsquark
Quote:

Originally Posted by elocin
I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!

Recall that the integration variable is a "dummy" variable.
$\int d \theta = \theta + C$

is the same process as
$\int dx = x + C$

-Dan
• Jun 21st 2008, 06:10 PM
Chris L T521
Quote:

Originally Posted by elocin
I'm not sure, I know I would use the sine triangle if I had tangent or cosine or something, but all I have is theta!

Well, we made the substitution $x=\sin(\theta)$. Solving for theta, we get $\sin^{-1}(x)=\sin^{-1}(\sin(\theta)) \implies \color{red}\boxed{\theta=\sin^{-1}(x)}$.

So instead of $\theta+k$ as our answer, it would be $\sin^{-1}(x)+k$.

Hope this makes sense! :D

--Chris
• Jun 21st 2008, 06:14 PM
elocin
So integration just gives you x? multiplied by 41? I'm confused!
• Jun 21st 2008, 06:15 PM
elocin
Oh okay!! Thank you!
• Jun 21st 2008, 06:20 PM
elocin
So I got 41arcsin(x), so would I just plug in 0 and 1? That gives me 41arcsin(1), since arcsin(0) =0.
• Jun 21st 2008, 06:31 PM
Chris L T521
Quote:

Originally Posted by elocin
So I got 41arcsin(x), so would I just plug in 0 and 1? That gives me 41arcsin(1), since arcsin(0) =0.

Yes. Note that $\sin^{-1}(1)=\frac{\pi}{2}$. So your answer is $42\bigg[\frac{\pi}{2}\bigg]=\color{red}\boxed{21\pi}$

Hope this makes sense! :D

--Chris
• Jun 21st 2008, 06:39 PM
elocin
Well the original integral was integrate from 0 to 1, the integral of (41) over (square root of 1-x^2), so wouldn't it be 41pi/2? I'm sorry, I just don't know where the 42 came from!
• Jun 21st 2008, 06:41 PM
Chris L T521
Quote:

Originally Posted by elocin
Well the original integral was integrate from 0 to 1, the integral of (41) over (square root of 1-x^2), so wouldn't it be 41pi/2? I'm sorry, I just don't know where the 42 came from!

It was a typo. (Headbang) It should be $\frac{41\pi}{2}$