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Homework Help: What is the definite integral of 1/(36+x^2) with bounds [0, 6]

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the definite integral of 1/(36+x^2) with bounds [0, 6]?

    I've only been taught U substitution to handle problems like these. I let u = 36+x^2 and du=2xdx but I am stuck and don't know what to do. Te answer is pi/24 but I don't know how to obtain it.

    Is there a better u to choose? if so, what?
     
  2. jcsd
  3. May 1, 2010 #2

    gabbagabbahey

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    Hint: What is [itex]\frac{d}{du}\tan^{-1}(u)[/itex]?
     
  4. May 1, 2010 #3
    U substitution isn't needed. Look at an integration table.

    Hint: The function being integrated takes the shape of 1/(x2+a2)
     
  5. May 1, 2010 #4
    Okay, I used that formula u gave me although in my book it is written as

    du/a^2+u^2 = 1/a*tan^-1(u/a)+C

    But I had to evaluate it with a calculator [0, 6]

    1/6*tan^-1(1)-1/6*tan^-1(0) = 0.1308 which is about pi/24

    How would I solve it without a calculator?
     
  6. May 1, 2010 #5
    Well, 1/6*tan^-1(1)-1/6*tan^-1(0)=1/6*tan^-1(1)-0=1/6*tan^-1(1). So I guess you would need to know that tan^-1(1)=[tex]\pi[/tex]/4.
     
  7. May 1, 2010 #6

    gabbagabbahey

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    Keep in mind that [itex]u=\tan^{-1}(x)[/itex] means that [itex]\tan(u)=\frac{\sin(u)}{\cos(u)}=x[/itex]...so where does [itex]\frac{\sin(u)}{\cos(u)}[/itex] equal zero? Where does it equal one?
     
  8. May 1, 2010 #7
    factor out a 36 in the denominator and in the denominator you'll get: 36(1+(x/6)^2)

    let u = x/6 and when you take the derivative and substitute in the original, you should get:

    (1/6)[1/(1 + u^2)] and integrate, remembering that 1/(1 + u^2) is tan^-1 u after you integrate, then you can go from there.
     
  9. May 3, 2010 #8
    answer after you integrate is (1/6)arctan(x/6) + C..

    then just evaluate it for 0 to 6

    (1/6)arctan(6/6)-[(1/6)arctan(0/6)]

    [(1/6)*(Pi/4)]-0= Pi/24
     
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