Results 1 to 2 of 2

Math Help - Big trigonometric substitution problem.

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    101

    Big trigonometric substitution problem.

    Okay, it's the last problem I have to do in the section and it stumped me.

    It's the integral of ((4-x)/x)^(1/2)dx

    The book says to let x=u^2.

    The answer is 4*arcsin(x^(1/2)/2) + x^(1/2) * (4-x)^(1/2) + C.

    No matter what I try I am not getting close to this answer. I know I'm supposed to get it into two pieces, one with form integral du/(a^2-u^2) .

    I start by letting u = x^(1/2). u^2=x. du= 1/2*x^(-1/2)

    So this gives me 2*integral square root of (4-u^2)/u^2 du.

    I then have to substitute. Since the integral is in the form of the square root of a^2-x^2 I use u=a*sin(t) or U=2*sin(t) and du=2*cos(t).

    This gives me the integral of 2*((4-4sin^2(t))/(4sin^2(t))^(1/2) * 2*cos(t).

    I think I have it right down to this point. Then when I retry the problem I keep trying different things from here on out.

    Sorry for the TI-89 format. I'm going to try to learn LaTex in the next week or so.

    Thanks for your help!
    Last edited by Wolvenmoon; February 19th 2011 at 05:43 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Wolvenmoon View Post
    Okay, it's the last problem I have to do in the section and it stumped me.

    It's the integral of ((4-x)/x)^(1/2)dx

    The book says to let x=u^2.

    The answer is 4*arcsin(x^(1/2)/2) + x^(1/2) * (4-x)^(1/2) + C.

    No matter what I try I am not getting close to this answer. I know I'm supposed to get it into two pieces, one with form integral du/(a^2-u^2) .

    I start by letting u = x^(1/2). u^2=x. du= 1/2*x^(-1/2)

    So this gives me 2*integral square root of (4-u^2)/u^2 du.

    I then have to substitute. Since the integral is in the form of the square root of a^2-x^2 I use u=a*sin(t) or U=2*sin(t) and du=2*cos(t).

    This gives me the integral of 2*((4-4sin^2(t))/(4sin^2(t))^(1/2) * 2*cos(t).

    I think I have it right down to this point. Then when I retry the problem I keep trying different things from here on out.

    Sorry for the TI-89 format. I'm going to try to learn LaTex in the next week or so.

    Thanks for your help!

    \displaystyle{\int\sqrt{\frac{4-x}{x}}\,dx\,,\,\,u^2=x\Longrightarrow 2u\,du=dx} , and

    \displaystyle{\int\sqrt{\frac{4-u^2}{u^2}}\,2u\,du=2\int\sqrt{4-u^2}\,du=4\int\sqrt{1-\left(\frac{u}{2}\right)^2}}\,du}

    Put now \displaystyle{\frac{u}{2}=\sin t\Longrightarrow du=2\cos t\,dt\Longrightarrow 4\int\sqrt{1-\left(\frac{u}{2}\right)^2}}\,du=8\int \cos^2t\,dt=}

    \displaystyle{=4(t+sin t\cos t)+C=4\left(\arcsin \frac{u}{2}+\frac{u}{2}\sqrt{1-\left(\frac{u}{2}\right)^2\right)+C and etc.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trigonometric substitution-another problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2011, 08:21 AM
  2. Trigonometric substitution problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 15th 2011, 02:34 PM
  3. Trigonometric substitution problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2010, 02:36 PM
  4. Trigonometric Substitution Problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 20th 2010, 08:05 PM
  5. Trigonometric Substitution Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 19th 2009, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum