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Integral of 1/(x^2 + 36)dx When and how to draw triangle?

  1. Mar 4, 2013 #1
    1. ∫1/(x2 + 36)dx

    2. I started by trying a trig substitution.

    The normal form, "a2 + x2, x=(a)tan(θ)," I thought could be reversed here:

    x2 + 62
    x = 6tan(θ)
    dx= 6sec2θdθ

    ∫1/[(6tanθ)2 + 36] = ∫1/[36(tan2θ + 1)]*6sec2θdθ

    = ∫1/[36sec2θ]*6sec2θdθ

    = ∫1/6dθ

    = (1/6)θ + C

    From before, θ= arctan(1/6(x))

    1/6(arctan(1/6(x))) + C

    In my class, sometimes the professor says we need to draw a triangle to figure out the value...

    I'm a little confused about that. Would I need to draw a triangle here to figure out an exact value?

    Since tan(1/6(x))=θ, do I draw a triangle and say the side opposite θ equals 1, and the adjacent side equals 6?

    And then do I figure out from that what arctanθ equals...?
    Would you help me with the triangle thing?

    Thank you so much! :D
  2. jcsd
  3. Mar 4, 2013 #2
    What value do you need to figure out?
  4. Sep 20, 2015 #3
    I used partial fractions... But it's a bit odd


    1=A(x+6i) + B(x-6i)
    A= -1/12i , B = 1/12i
    1/i= -i
    i/12 Integ 1/(x+6i) -1/(x-6i) dx

    i/12 ln |(x+6i)/(x-6i)| + c
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