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Homework Help: Having trouble with Integration by Parts

  1. Aug 23, 2004 #1
    I'm real stuck with this problem of mine in The Calculus 7 by Leithold

    \int arctan \sqrt{x} dx

    Since there is no elementary formula for integration of an inverse trigo function, we cannot manipulate the integrand in such a way as to integrate easily with one step of Integration by parts.(please verify)

    So first, If we let [tex]u = arctan \sqrt{x}[/tex] and
    [tex]dv = dx[/tex]


    [tex]du = \frac{dx}{2 \sqrt{x} (x+1)}[/tex] and

    then we have

    \int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}

    Now, Case 1, without simplifying [tex]\frac{x}{\sqrt {x}}[/tex]

    let [tex]db = \frac{1}{2 \sqrt{x} (x+1)}dx[/tex]
    and [tex]a = x[/tex].
    Then [tex] da = dx[/tex]
    and [tex] b = arctan \sqrt{x} [/tex]


    \int arctan \sqrt{x} dx = xarctan \sqrt{x} - xarctan \sqrt{x} + \int arctan \sqrt{x} dx [/tex]

    which of course will lead us nowhere but 0=0

    Now, Case 2 - If we simplify [tex]\frac{x}{\sqrt {x}} = \sqrt{x}[/tex]

    let [tex]a= \sqrt{x}[/tex] and [tex] db = \frac{dx}{1+x} [/tex],
    then [tex]da = \frac{dx}{2 \sqrt{x}}[/tex] and [tex] b = ln (x+1)[/tex]


    \int arctan \sqrt{x} dx = xarctan \sqrt{x} - \frac{1}{2} ( \sqrt{x}ln (x+1) - \int \frac{ln |1+x| dx}{2 \sqrt{x}}) [/tex]

    and if we do Integration by parts again, we will just get an infinite sequence of [tex] \sqrt{x}ln (x+1) [/tex] (please verify)

    I'm really open to suggestions ^_^`
  2. jcsd
  3. Aug 23, 2004 #2
  4. Aug 23, 2004 #3
    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}[/tex]

    From there... add 1 and -1 to your numerator, and disect the expression into two... hope you figure out the next steps. (in your integral of udv)
  5. Aug 23, 2004 #4
    0.0 hey... why didn't I think of that???

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}[/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{x + 1 - 1}{2 \sqrt{x} (x+1)}dx[/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{(x + 1)}{2 \sqrt{x} (x+1)} - \frac{1}{2 \sqrt{x} (x+1)}) dx[/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{1}{2 \sqrt{x}} - \frac{1}{2 \sqrt{x} (x+1)}) dx[/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{dx}{2 \sqrt{x}} + \int \frac{dx}{2 \sqrt{x} (x+1)}[/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{dx}{2 \sqrt{x} (x+1)})[/tex]

    let [tex]u^2 = x[/tex], then [tex]u = \sqrt{x}[/tex], since [tex]du = \frac{1}{2 \sqrt{x}} [/tex]it can be integrated directly and it yields arctan...


    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{du}{(u^2+1)} [/tex]

    [tex]\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + arctan \sqrt{x} + C[/tex]

    Is this correct? ^_^;;

    Man... I really don't like it when the answer is just right in your face and slaps you XD. Gotta remember the +a -a concept ^_^;;; Thanx btw Irony of Truth
  6. Aug 23, 2004 #5
    now that's correct.
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