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

1. ## 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!
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P.S. I attached the picture of the problem so that noone gets confused.

2. Hi

Through $u = 4x$

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

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

3. Originally Posted by running-gag
Hi

Through $u = 4x$

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

Then you can use $u = \sin \theta$
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 $\theta$ 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.
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4. 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)

Originally Posted by starbless
Also, how do I use sin $\theta$ to solve the rest of the problem?
$u = \sin \theta$ and u varies from -1 to 1 therefore $\theta$ varies from $-\frac{\pi}{2}$ to $+\frac{\pi}{2}$
And $du = \cos \theta d\theta$

5. You should notice that this is an improper integral, it should be solved in terms of limits ..