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Homework Help: Trig integral

  1. Nov 8, 2006 #1
    I'm in the middle of solving [tex]\int\sqrt{x^2+9}dx[/tex] and I got it into the form of [tex]3\int\sec^3\theta d\theta[/tex], and I'm pulling a blank. Where does one begin? I could integrate by parts, setting u=sec(theta) and dv=sec^2(theta), but it's not getting me very far. Any suggestions?
     
  2. jcsd
  3. Nov 8, 2006 #2
    bummer. i had that and decided it didn't look appealing...

    OK, how's this?

    [tex]3\int\sec^3\theta d\theta[/tex]

    =[tex]3(\sec\theta\tan\theta-\int\tan^2\theta\sec\theta d\theta)[/tex]

    =[tex]3(\sec\theta\tan\theta-\int\frac{sin^2\theta}{cos^3\theta}d\theta)[/tex]

    =[tex]3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)[/tex]

    =[tex]3(\sec\theta\tan\theta-\int\frac{1+\cos^2\theta}{cos^3\theta}d\theta)[/tex]

    =[tex]3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)[/tex]

    Let I=[tex]\int\sec^3\theta d\theta[/tex]

    Then [tex]3I=3(\sec\theta\tan\theta-\int\sec^3\theta d\theta +\int\sec\theta d\theta)[/tex]

    The 3's cancel, so [tex]I+I=(\sec\theta\tan\theta+\int\sec\theta d\theta)[/tex]

    so [tex]I=(\sec\theta\tan\theta+\int\sec\theta d\theta)[/tex] and I can evaluate it from there (I hope).

    Thank you very much!

    Where did your post go, Courtigrad? I used it!
     
    Last edited: Nov 8, 2006
  4. Nov 9, 2006 #3

    dextercioby

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    Try the substitution [itex] x=3\sinh t [/itex].

    Daniel.
     
  5. Nov 9, 2006 #4

    HallsofIvy

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    Here's how I would almost automatically do an integral like that. First convert secant to cosine
    [tex]\int sec^3 x dx= \int \frac{1}{cos^3(x)}dx[/itex]
    which is an odd power of cosine. "Take out" a cos(x) to use with dx. That is, multiply both numerator and denominator by cos(x)
    [tex]\int \frac{cos (x)}{cos^4(x)}dx= \int \frac{cos(x)dx}{(1- sin^2(x))^2}[/tex]
    Now let u= sin(x) so du= cos(x)dx
    [tex]\int \frac{du}{(1- u^2)^2}= \int \frac{du}{(1-u)^2(1+u)^2}[/tex]
    and now use partial fractions.
     
  6. Nov 9, 2006 #5
    Here's how I integrated the [tex]\int sec^3(x)dx[/tex] term.

    [tex]\int sec^3(x)dx[/tex]
    [tex]=\int sec^2(x)sec(x)[/tex]

    [tex]u = sec(x)[/tex]
    [tex]du = sec(x)tan(x) dx[/tex]

    [tex]dv = sec(x)^2(x)dx[/tex]
    [tex]v = tan(x) [/tex]


    [tex]\int sec^3(x)dx[/tex]
    [tex]= sec(x)tan(x) - \int tan(x)sec(x)tan(x) dx [/tex]
    [tex]= sec(x)tan(x) - \int tan^2(x)sec(x) dx [/tex]
    [tex]= sec(x)tan(x) - \int (sec^2(x) - 1)sec(x) dx [/tex]
    [tex]= sec(x)tan(x) - \int (sec^3(x) - sec(x) dx [/tex]
    [tex]= sec(x)tan(x) - \int (sec^3(x) + \int sec(x) dx [/tex]

    Now we have [tex]\int sec^3(x) [/tex] on both sides, so consolidate them.

    [tex] 2 \int sec^3(x) dx = sec(x)tan(x) + \int sec(x) dx [/tex]
    [tex] 2 \int sec^3(x) dx = sec(x)tan(x) + ln|sec(x) + tan(x)| + K [/tex]
    [tex] \int sec^3(x) dx = {{sec(x)tan(x) + ln|sec(x) + tan(x)|}\over{2}} + C [/tex]


    How did I get
    [tex] \int sec(x) dx = ln|sec(x) + tan(x)|[/tex]?

    Well, simplify this integral:

    [tex] \int sec(x) dx = \int sec(x)*{{sec(x)+tan(x)}\over{sec(x)+tan(x)}}[/tex]
     
    Last edited: Nov 9, 2006
  7. Nov 9, 2006 #6
    i would have used the substitution x = 3 tan(t)
     
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