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Integrals with trig substitutions =p

  1. Dec 5, 2009 #1
    NVM I GET HOW TO DO I TNOW
    1. The problem statement, all variables and given/known data

    "The region bounded by the graphs of [tex]y = \frac{x}{\sqrt{x^2+25}} [/tex], y = 0, and x = 5 is revolved about the y-axis. Find the volume of the resulting solid."


    2. Relevant equations

    The only way i see to do this right now is to use shells, and the equation for that is

    [tex] V = \int 2 \pi x f(x) [/tex] where the integral is from the lower limit to the upper limit
    Mathematica cant find an answer to it. You can try it yourself if you want [Integral of 2*pi*x^2/(x^2+25)^(1/2)]


    Thanks in advance for your help


    Oh also, the correct answer is [tex] 25\pi[\sqrt{2}-\ln{\sqrt{2}+1}] \approx 41.85 [/tex]



    EDIt: ok mathematica does get an answer, but is the a way to find it without a reduction formula for tanx^2*secx ?
     
    Last edited: Dec 5, 2009
  2. jcsd
  3. Dec 5, 2009 #2

    LCKurtz

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    I get an integrand containing

    [tex]\frac{\tan^2\theta}{\sec\theta}= \frac{\sec^2\theta - 1}{\sec\theta}=\sec\theta-\cos\theta[/tex]

    Do you know how to integrate [itex]\sec\theta[/itex]?
     
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