Results 1 to 9 of 9

Math Help - Integral using given substitute...

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    66

    Integral using given substitute...

    Hi there, I'm having some difficulty with the following problem:

    Integrate: \int \frac{\sqrt{t-1}}{t-2}dt using u= \sqrt{t-1}

    I've found that the denominator = u^2-1 and


    \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u} and therefore dt=2u\ du

    But am not sure if I'm on the right track and where to go from there, because any progress I've made doesn't give me the answer in the back of the book.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,897
    Thanks
    327
    Awards
    1
    Quote Originally Posted by flashylightsmeow View Post
    Hi there, I'm having some difficulty with the following problem:

    Integrate: \int \frac{\sqrt{t-1}}{t-2}dt using u= \sqrt{t-1}

    I've found that the denominator = u^2-1 and


    \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u} and therefore dt=2u\ du

    But am not sure if I'm on the right track and where to go from there, because any progress I've made doesn't give me the answer in the back of the book.
    \displaystyle \int \frac{\sqrt{t - 1}}{t - 2}~dt = \int \frac{2u^2}{u^2 - 1}~du which can be evaluated using a number of different methods. What answer are you getting?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    Quote Originally Posted by flashylightsmeow View Post
    Integrate: \int \frac{\sqrt{t-1}}{t-2}dt using u= \sqrt{t-1}

    I've found that the denominator = u^2-1 and


    \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u} and therefore dt=2u\ du
    Correct till this. Then \int\frac{u}{u^2-1}(2u.du) = \int 2\frac{u^2}{u^2-1}du

    = 2\int (1+\frac{1}{u^2-1})

    = 2(u+\frac{1}{2}\log\frac{1-u}{1+u}) + constant

    Substitute back u=\sqrt{t-1}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    Quote Originally Posted by flashylightsmeow View Post
    Integrate: \int \frac{\sqrt{t-1}}{t-2}dt using u= \sqrt{t-1}

    I've found that the denominator = u^2-1 and


    \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u} and therefore dt=2u\ du
    Correct till this. Then \int\frac{u}{u^2-1}(2u.du) = \int 2\frac{u^2}{u^2-1}du

    = 2\int (1+\frac{1}{u^2-1})

    = 2(u+\frac{1}{2}\log\frac{1-u}{1+u}) + constant

    Substitute back u=\sqrt{t-1}.


    EDIT: ooops....sorry
    Last edited by Sambit; April 3rd 2011 at 10:22 AM. Reason: sorry for double posting
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2011
    Posts
    66
    Ah ok, I got to this bit, and took the 2 outside the integral, but how do you split the integrand into partial fractions?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,897
    Thanks
    327
    Awards
    1
    Quote Originally Posted by flashylightsmeow View Post
    Ah ok, I got to this bit, and took the 2 outside the integral, but how do you split the integrand into partial fractions?
    \displaystyle \frac{1}{u^2 - 1} = \frac{A}{u + 1} + \frac{B}{u - 1}

    Thus
    A(u - 1) + B(u + 1) = 1 \implies A + B = 0 \text{ and } -A + B = 1

    What are A and B?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2011
    Posts
    66
    so A=-1/2 and B=1/2 so \int \frac{1}{2(u-1)}+\frac{1}{2(u+1)} but how do I get from this to Sambit's integral above?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,897
    Thanks
    327
    Awards
    1
    Quote Originally Posted by flashylightsmeow View Post
    so A=-1/2 and B=1/2 so \int \frac{1}{2(u-1)}+\frac{1}{2(u+1)} but how do I get from this to Sambit's integral above?
    \displaystyle \int\frac{2u^2}{u^2-1}~du = 2 \int \left ( 1 + \frac{1}{u^2-1} \right )~du

    \displaystyle = 2\int du + 2 \int \frac{1}{u^2-1}du = 2\int du + 2 \int \left ( \frac{-1/2}{u + 1} + \frac{1/2}{u - 1} \right )  ~du

    \displaystyle = 2\int du - \frac{2}{2} \int \frac{1}{u + 1} + \frac{2}{2} \int \frac{1}{u - 1}~du

    Can you take it from here?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2011
    Posts
    66
    Got it! Thank you very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Substitute suitable value of y to find 5^{1/3}
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 27th 2011, 09:22 PM
  2. integrate using substitute
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 19th 2008, 05:58 AM
  3. Replies: 1
    Last Post: October 15th 2008, 06:28 AM
  4. Integration by substitute problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 31st 2007, 03:05 AM
  5. Replies: 2
    Last Post: June 26th 2007, 01:13 AM

Search Tags


/mathhelpforum @mathhelpforum