Results 1 to 5 of 5

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

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    9

    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.
    Attached Thumbnails Attached Thumbnails integral between 1/4 and -1/4 of (1/(sqrt(1-(4x)^2))dx-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    9
    Quote Originally Posted by running-gag View Post
    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.
    -star
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by starbless View Post
    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 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    You should notice that this is an improper integral, it should be solved in terms of limits ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral of sqrt(1+cos x)
    Posted in the Calculus Forum
    Replies: 11
    Last Post: June 4th 2010, 10:23 AM
  2. Integral of sin^3 (sqrt(x)) / (sqrt(x))
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 21st 2010, 05:48 PM
  3. Integral of sqrt(9-x^2)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 18th 2009, 10:34 PM
  4. Replies: 0
    Last Post: September 10th 2008, 08:53 PM
  5. Replies: 11
    Last Post: January 6th 2008, 10:33 AM

Search Tags


/mathhelpforum @mathhelpforum