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Trig subtitution

  1. Nov 8, 2003 #1
    Can someone plz show me the concept of trig substituion and easy way to use it...I don't quite understand this...
    I'm ot sure how to dothis problem below..
    1) indefinite integral dx/(x^3sqrt(x^2-a^2))
    2) indefinite integral (1-4x^2)^(3/2)dx

    How to use integration by part for
    indefinite integral of x*inverse tan(x) dx.
  2. jcsd
  3. Nov 8, 2003 #2
    #1: It is the ∫ Sec-1x * x-2 dx.

    Use parts formula: let u = Sec-1, du = x'/|x|sqrt(x^2 - a^2) where a = 1. dv = x-2 dx, v = -x^-1.

    The answer? I'm not sure, but out of scratch i got -sec^-1x/x + 2sqrt(x^6 - x^4)/6x^5 - 4x^3 + C = -Cosx/x - sinx * x^3/3
    Last edited: Nov 8, 2003
  4. Nov 9, 2003 #3


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    The point of "trigonometric substitution" is to use trig identities such as
    sin2(θ)= 1- cos2(θ) so that
    sin(θ)= √(1- cos2(θ))

    sec2(θ)= 1+ tan2(θ) so that
    sec(θ)= √(1+ tan2(θ))

    tan2(θ)= sec2(θ)- 1 so that
    tan(θ)= √(sec2(θ)- 1)

    When you see something like ∫dx/(x3√(x2-a2)) you should immediately think "Hmmm, that √(x2- a2) reminds me of √(sec2(θ)- 1)- especially if I factor out a2 to get
    a√((x/a)2-1). I'll bet making a substitution like
    x/a= sec(θ) (0r x= a sec(θ)) will work!

    Doing that, √(x2- a2) becomes √(a2sec2(θ)- a2)= a√(sec2-1)= a√(tan2)= a tan(θ)!

    Also, x3 become a3sec3(θ) and dx= a d(sec(θ))= a sec(θ)tan(θ) dθ

    The entire integral becomes ∫(a sec(θ)tan(θ)dθ/(a3sec3(θ)(a tan(θ)))
    = (1/a3)∫(1/sec2(θ)
    = (1/a3)∫cos2(θ)dθ
    which can be done by using the trig identity:
    cos2(θ)= (1/2)(1+ cos(2θ)).
  5. Nov 9, 2003 #4
    i was thinking about that and actually shouldn't it be |atanθ|, not atanθ? the a vs |a| part doesn't matter because with the x3, you'd have a3|a|=a4 anyway but there'd still be a |tanθ|. perhaps when the integration is done and you go back to x, it won't matter if it was tanθ or |tanθ|...
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