Results 1 to 5 of 5

Math Help - Integrals of Trig Functions

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Integrals of Trig Functions

    I'm having trouble with this problem:

    tan^2(x^1/2)/x^(1/2)
    I don't know how to get any special characters in there so just assume I want to integrate the above problem. Just for clarification...it reads tan squared square root x. (The entire tan function is squared) divided by square root x.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by nick898 View Post
    I'm having trouble with this problem:



    I don't know how to get any special characters in there so just assume I want to integrate the above problem. Just for clarification...it reads tan squared square root x. (The entire tan function is squared) divided by square root x.
    \int \frac{tan^2(\sqrt{x})}{\sqrt{x}}~dx

    Let y = \sqrt{x} \implies dx = 2y~dy

    So
    \int \frac{tan^2(\sqrt{x})}{\sqrt{x}}~dx = \int \frac{tan^2(y)}{y} \cdot 2y~dy

    = 2 \int tan^2(y)~dy

    How 'bout now?

    -Dan
    Last edited by topsquark; December 4th 2007 at 12:24 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    10
    Ok. I understand how to get the answer out of that. (It has the answers in the back of the book) but I'm confused how you got dx = -2ydy

    I think it's just a minor difference in the way I was taught but I just can't see it.

    I understand using a variable to represent square root x. I just don't get the dx = -2ydy part.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by nick898 View Post
    Ok. I understand how to get the answer out of that. (It has the answers in the back of the book) but I'm confused how you got dx = -2ydy

    I think it's just a minor difference in the way I was taught but I just can't see it.

    I understand using a variable to represent square root x. I just don't get the dx = -2ydy part.
    Sorry, I was off by a - sign:
    y = \sqrt{x}

    dy = \frac{1}{2\sqrt{x}}~dx

    dy = \frac{1}{2y}~dx

    dx = 2y~dy

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2007
    Posts
    10
    Ok, it all makes sense now. Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 4th 2011, 09:17 PM
  2. [SOLVED] Integrals Involving Powers of Trig Functions.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 28th 2010, 01:26 AM
  3. Replies: 1
    Last Post: February 3rd 2010, 03:25 PM
  4. Integrals Involving Inverse Trig Functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 12th 2009, 10:51 PM
  5. Evaluating Integrals (inverse trig functions)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 22nd 2009, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum