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Homework Help: Integral involving trig

  1. Jul 21, 2006 #1
    [tex] \int \frac{dx}{x\sqrt{a^{2}+x^{2}}} [/tex].

    So, [tex] x = a\tan\theta [/tex], and [tex] dx = a\sec^{2}\thetha d\theta [/tex]. When we substitute we get: [tex] \int\frac{a\sec^{2}\theta}{(a\tan\theta)(a\sec\theta}) [/tex] which equals [tex] \frac{1}{a}\int \csc \theta d\theta [/tex]. I know that [tex] \int \csc \theta d\theta = -\ln|\csc\theta + \cot\theta| [/tex]. And [tex] \theta = \tan^{-1}(\frac{x}{a}) [/tex]. So I substitute this into the equation. How do we get from that to this:

    [tex] (\frac{1}{a})\ln|\frac{x}{a+\sqrt{a^{2}+x^{2}}} [/tex]

    Last edited: Jul 21, 2006
  2. jcsd
  3. Jul 21, 2006 #2


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    Can you write cot([itex]\theta[/itex]) and csc([itex]\theta[/itex]) in terms of tan([itex]\theta[/itex])?
  4. Jul 21, 2006 #3
    yeah and I got [tex] \frac{\cos\theta}{\sin\theta}+\frac{1}{\sin\theta} = \frac{1+\cos\theta}{\sin\theta} [/tex]
  5. Jul 21, 2006 #4


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    Not in terms of [itex]\theta[/itex], in terms of tan([itex]\theta[/itex]). For example, cot([itex]\theta[/itex])=1/tan([itex]\theta[/itex]). What is csc([itex]\theta[/itex])? (hint: use sec2([itex]\theta[/itex])=1+tan2([itex]\theta[/itex]) )
  6. Jul 21, 2006 #5
    [tex]-ln(\frac{1}({\tan\theta}+\sqrt{1+(\frac{1}{\tan\theta})^{2}) [/tex]
  7. Jul 21, 2006 #6
    i got it. thanks
  8. Jul 21, 2006 #7


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    [tex]x=a\tan \theta[/tex] gives [tex]\tan \theta = \frac{x}{a}[/tex] so we have a right triangle where: the length of the leg oppsite [tex]\theta[/tex] is x, the length of the leg adjacent to [tex]\theta[/tex] is a, and the hypotenuse is [tex]\sqrt{a^2+x^2}.[/tex] From the triangle it follows that [tex]\cot \theta = \frac{a}{x},[/tex] what is [tex]\csc \theta[/tex]?
  9. Jul 21, 2006 #8


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    An alternative way of solving it is to let
    [tex]u = \sqrt{x^2 + a^2}[/tex]
    [tex]du = \frac{x}{\sqrt{x^2 + a^2}}dx[/tex]
    Then the integral becomes
    [tex]\int \frac{du}{u^2-a^2}[/tex]
    which can be solved by partial fraction decomposition.
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