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Math Help - integration by trig substitution giving me different, equivalent answers, as calcula?

  1. #1
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    integration by trig substitution giving me different, equivalent answers, as calcula?

    Hi. Using trig sub to get integrals, I keep getting different but possibly equivalent answers as in book, and on calculator.

    For example, integral(5/sqrt(25x^2-9),x)

    By trig sub,

    I get: ln|5x/3+(5sqrt(x^2-9/25)/3|+c

    Calculator gets: ln(sqrt(25x^2-9)+5x)

    Wolfram integrator gets: log(2(sqrt(25x^2-9)+5x))

    But, I am pretty sure I did the problem correctly.

    Are they equivalent? If they are, How can I show that they are to avoid my teacher marking off points?
    Last edited by StudentMCCS; March 10th 2012 at 09:02 PM.
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  2. #2
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    Re: integration by trig substitution giving me different, equivalent answers, as calc

    Hint :

    Substitute ~x=\frac{3}{5\cdot \cos u}

    Result should be :

     I= \ln (\sqrt{25x^2-9}+5x) + C
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  3. #3
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    Re: integration by trig substitution giving me different, equivalent answers, as calc

    Quote Originally Posted by princeps View Post
    Hint :

    Substitute ~x=\frac{3}{5\cdot \cos u}

    Result should be :

     I= \ln (\sqrt{25x^2-9}+5x) + C
    Thats what I did.

    The integral simplifies first to integral(1/sqrt(x^2-9/25),x)

    x=(3/5)*sec(u)
    ...

    In the end it is integral(sec(u),u)

    =ln|sec(u)+tan(u)|+c

    I have a right triangle, where hypotenuse is x, opposite is sqrt(x^2-9/25), and adjacent=3/5.

    So, sec(u)=1/((3/5)/x)=5x/3, and tan(u)=sqrt(x^2-9/25)/(3/5)=sqrt(25x^2-9)/3.

    So I get:  I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C
    Last edited by StudentMCCS; March 10th 2012 at 11:14 PM.
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  4. #4
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    Re: integration by trig substitution giving me different, equivalent answers, as calc

    Quote Originally Posted by StudentMCCS View Post
    Thats what I did.

    The integral simplifies first to integral(1/sqrt(x^2-9/25),x)

    x=(3/5)*sec(u)
    ...

    In the end it is integral(sec(u),u)

    =ln|sec(u)+tan(u)|+c

    I have a right triangle, where hypotenuse is x, opposite is sqrt(x^2-9/25), and adjacent=3/5.

    So, sec(u)=1/((3/5)/x)=5x/3, and tan(u)=sqrt(x^2-9/25)/(3/5)=(5*sqrt(25x^2-9))/3.

    So I get:  I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C
    You did everything ok .

     I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C_1=\ln (\sqrt{25x^2-9}+5x)+ \ln \frac{1}{3}+ C_1 ~ , hence :

     I= \ln (\sqrt{25x^2-9}+5x)+ C
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  5. #5
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    Re: integration by trig substitution giving me different, equivalent answers, as calc

    Quote Originally Posted by princeps View Post
    You did everything ok .

     I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C_1=\ln (\sqrt{25x^2-9}+5x)+ \ln \frac{1}{3}+ C_1 ~ , hence :

     I= \ln (\sqrt{25x^2-9}+5x)+ C
    Ahh, that makes sense. Thanks. I wonder which way my teacher would prefer I wrote it?
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