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Trigonometric Substitution problem

  1. Sep 14, 2013 #1
    1. [itex]∫\frac{\sqrt{x^{2}-4}}{x} dx[/itex], [itex]x=2secθ[/itex], [itex]dx=2secθtanθ dθ[/itex]

    2. [itex]\sqrt{x^{2}-a^{2}}[/itex],[itex]sec^{2}θ-1=tan^{2}θ[/itex]

    3. [itex]\sqrt{x^{2}-4}=\sqrt{4sec^{2}θ-4}=\sqrt{4(1+tan^{2}θ)-4}=\sqrt{4tan^{2}θ}=2\left|tanθ\right|=2tanθ[/itex];[itex]∫\frac{\sqrt{x^{2}-4}}{x}dx=∫\frac{2tanθ}{2secθ}dθ=\frac{ln\left|secθ\right|}{ln\left|secθ+tanθ\right|}+C=ln\left|secθ\right|-ln\left|secθ+tanθ\right|+C[/itex];[itex]∫\frac{\sqrt{x^{2}-4}}{x}dx=ln\left|\frac{x}{2}\right|-ln\left|\frac{x}{2}+\frac{\sqrt{x^{2}-4}}{2}\right|+C=ln\left|x\right|-ln\left|2\right|-ln\left|x+\sqrt{x^{2}-4}\right|+ln\left|2\right|+C[/itex]. Please tell me what I did wrong.
  2. jcsd
  3. Sep 14, 2013 #2


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    What happened to the substitution for dx in terms of dθ?
  4. Sep 14, 2013 #3
    I see what you mean. Ok, here is what I have after the dx substitution: [itex]\int\frac{\sqrt{x^{2}-4}}{x}dx = \int\frac{2tanθ}{2secθ}2secθtanθdθ = 2\int tan^{2}θdθ[/itex]
  5. Sep 14, 2013 #4
    Which is: [itex]2\int(sec^{2}θ-1)dθ = 2(tanθ-θ)+C[/itex]
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