# integral between 1/4 and -1/4 of (1/(sqrt(1-(4x)^2))dx

• Mar 17th 2010, 01:23 PM
starbless
integral between 1/4 and -1/4 of (1/(sqrt(1-(4x)^2))dx
Hi. I missed today's class and was wondering if someone could assist me with solving this problem.

Thanks!
-star

P.S. I attached the picture of the problem so that noone gets confused.
• Mar 17th 2010, 01:28 PM
running-gag
Hi

Through $\displaystyle u = 4x$

$\displaystyle \int_{-\frac14}^{\frac14} \frac{dx}{\sqrt{1-(4x)^2}} = \frac14 \int_{-1}^{1} \frac{du}{\sqrt{1-u^2}}$

Then you can use $\displaystyle u = \sin \theta$
• Mar 17th 2010, 01:37 PM
starbless
Quote:

Originally Posted by running-gag
Hi

Through $\displaystyle u = 4x$

$\displaystyle \int_{-\frac14}^{\frac14} \frac{dx}{\sqrt{1-(4x)^2}} = \frac14 \int_{-1}^{1} \frac{du}{\sqrt{1-u^2}}$

Then you can use $\displaystyle u = \sin \theta$

Okay, I've lost you. I see that you've brought a 1/4 to the outside by squaring 4x^2, but how come the range of the integral also changes?

Also, how do I use sin$\displaystyle \theta$ to solve the rest of the problem?

I don't mean for you to simply solve the problem for me; I simply want to follow how you're getting to that point.

Thanks so much.
-star
• Mar 17th 2010, 01:49 PM
running-gag
Quote:

Originally Posted by starbless
Okay, I've lost you. I see that you've brought a 1/4 to the outside by squaring 4x^2, but how come the range of the integral also changes?

x varies from -1/4 to +1/4
Therefore u=4x varies from -1 to+1
And du = 4 dx => dx = du / 4 (this is where 1/4 comes from)

Quote:

Originally Posted by starbless
Also, how do I use sin$\displaystyle \theta$ to solve the rest of the problem?

$\displaystyle u = \sin \theta$ and u varies from -1 to 1 therefore $\displaystyle \theta$ varies from $\displaystyle -\frac{\pi}{2}$ to $\displaystyle +\frac{\pi}{2}$
And $\displaystyle du = \cos \theta d\theta$
• Mar 17th 2010, 02:24 PM
General
You should notice that this is an improper integral, it should be solved in terms of limits ..