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Trig Substitution Insanity

  1. Oct 5, 2013 #1
    (4x2 + 9) 3/2

    According to my book this is a trig substitution integral. The normal procedure is to substitute atanθ for x when one has a square root w an argument of the form x^2 + a^2. Because the argument of the square root is 4x2 + 9, as opposed to simply x2 +9, the book suggests substituting 2x for x, hence instead of substituting 3tanθ for x- you sub 3/2tanθ. That much makes sense to me. The part that is driving me insane is that the book goes on to complete the trig substitution by substituting this x value (3/2 tanθ) for x in the numerator (ok), substituting 3/2sec2θdθ for dx (ok)... however in the denominator, they write √(4x^2 +9) = √9tan^2θ +9 = 3secθ. That's not substituting the "agreed upon" value of x (i.e., 3/2tanθ) it's just substituting 3tanθ! This is my first post and I'm no math genius so I'm certain I'm just not comprehending something but I've been stuck on this for a while and it's driving me nuts! Any help would greatly appreciated- I hope this post is fairly intelligible.
  2. jcsd
  3. Oct 5, 2013 #2
    Hey Grace, don't worry the post is legible - just plug the substitution into the denominator to see they are substituting in the agreed upon value for x:

    [itex] \sqrt{4x^2+9} = \sqrt{4(\tfrac{3}{2}\tan(\theta))^2+9} = [/itex]
    [itex]\sqrt{4(\tfrac{9}{4}\tan^2(\theta)) +9} = \sqrt{9 \tan^2(\theta) +9} = \sqrt{9 (\tan^2(\theta) +1)} = 3 \sqrt{\tan^2(\theta) +1} = 3 \sqrt{\sec^2(\theta)} = 3\sec (\theta)[/itex]

    I think you just forgot to write a [itex]4[/itex] in one step & got confused as a result of this. Also, this & this link show you that you don't have to use trig substitutions in this case necessarily, if you're interested :smile:
    Last edited by a moderator: Oct 5, 2013
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