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Use substitution x=2tan(y) to integrate (37dx)/(x^2*sqrt(x^2+4)) in y

  1. Feb 15, 2006 #1
    I feel so :cry: doing this problem.
    PLEASE HELP!!!! AND TEACH ME How TO DO IT.
    My question is this.

    use the substitution x=2tan(y)

    (37dx)/(x^2*sqrt(x^2+4))

    give the answer in terms of y.

    I did the substition, but it looked more complicated.
    It doesn't look like a u*du thing, but it does look like some arcsin thing... i couldn't figure it out. The x^2 is bugging me.
    PLease help!!!
     
  2. jcsd
  3. Feb 15, 2006 #2

    benorin

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    Consider that

    [tex]\sqrt{4+x^2}=\sqrt{4+4\tan^2(y)}=2\sqrt{1+\tan^2(y)}[/tex]

    and then use the approiate Pythagorean identity.
     
  4. Feb 15, 2006 #3

    benorin

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    namely [tex]1+\tan^2(y)=\sec^2(y)[/tex] so that

    [tex]2\sqrt{1+\tan^2(y)}=2\sqrt{\sec^2(y)}=2\sec (y)[/tex]
     
  5. Feb 15, 2006 #4

    benorin

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    Also, for [tex]x=2\tan (y)[/tex] we have [tex]dx=2\sec^2 (y)dy[/tex]
     
  6. Feb 15, 2006 #5
    i got

    original expression= (37/4)*((sec(y))/((sec(y))^2-1)*dy

    but I am stuck.

    I tried to do it so that its integral is ln(u)*du, but then sec's derivative would come up...
     
    Last edited: Feb 15, 2006
  7. Feb 15, 2006 #6

    HallsofIvy

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    I don't see how you could have gotten that!
    Your integral is
    [tex]\int\frac{37dx}{x^2\sqrt{x^2+4}}[/tex]
    Letting x= 2 tan y, gives,as you were told before, dx= 2 sec2y dy, and [itex]\sqrt{x^2+4}= \sqrt{4tan^2y+4}= 2sec y[/itex]. Of course x2= 4tan2 y. I don't see how you could have gotten that "-1" in "sec2y- 1".
     
  8. Feb 15, 2006 #7
    1+(tany)^2=(secy)^2
    (tany)^2=(secy)^2-1

    4(tany)^2=4((secy)^2-1)

    this is how i got it... what's wrong?
     
  9. Feb 16, 2006 #8

    HallsofIvy

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    Do you mean you changed that x2 inthe denominator into
    4tan2 y and then into 4(sec2y-1)?? Why?

    As was point out before, if x= 2 tan y, then dx= 2 sec2 y
    and x2+ 4= 4tan2y+ 4= 4 sec2y so that [itex]\sqrt{x^2y+ 4}= 2 sec y[/itex]. Your integral becomes
    [tex]37\int \frac{2 sec^2 y dy}{(4tan^2 y)(2 sec y)}[/tex]
    which reduces to
    [tex]\frac{37}{4}\int\frac{sec y dy}{tan^2 y}[/tex]
    Because of the odd power of sec y, I would be inclined to convert to sine and cosine now.
     
  10. Feb 17, 2006 #9
    thanx for the guide!!! I did it.
     
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