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Trigonometric substitution

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data

    evaluate the definite integral ∫(0 to 3) dx/sqrt(25+x^2)

    2. Relevant equations



    3. The attempt at a solution

    I first used substitution and set x=5tanθ, and dx=5tanθsecθdθ

    then i wrote the integral as 5∫ tanθsecθdθ/sqrt(25(1+tan^2(θ))

    after some simplification i got

    ∫tanθsecθdθ/secθ = ∫tanθ =tanθsecθ

    I then used θ=arctan(x/5) from the original x substitution and my final solution looked like this

    tan(arctan(3/5))sec(arctan(3/5))-tan(arctan(0))sec(arctan(0))

    It is incorrect, can anyone let me know where i might have done something incorrectly? Thanks
     
  2. jcsd
  3. Feb 7, 2013 #2

    jedishrfu

    Staff: Mentor

    if x=5 tan(theta) then isn't dx=5sec^2(theta) dtheta
     
  4. Feb 8, 2013 #3
    oh you're right, but after solving the integral i still end up with tanθsecθ.. Unless i am doing something wrong in finding my θ values for x=3 and x=0?
     
  5. Feb 8, 2013 #4

    Dick

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    No, you don't get tanθsecθ. And the integral of tanθ isn't even tanθsecθ. What are you doing?
     
  6. Feb 8, 2013 #5

    SammyS

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    If you use your substitution to find [itex]\displaystyle \ \int \frac{dx}{\sqrt{25+x*2}}\,,\ [/itex] you get [itex]\displaystyle \ \int\sec(\theta)\,d\theta\ .[/itex]

    The anti-derivative of sec(θ) is not tan(θ)sec(θ) . The derivative of sec(θ) is tan(θ)sec(θ) . You seem to have these confused.

    A better substitution uses the hyperbolic sine, x = sinh(u) .
     
  7. Feb 8, 2013 #6
    oh right thanks, i used ∫secx=ln(secx+tanx) and it worked. Just a question about the hyperbolic sine, would you mind explaining it a bit to me or linking me to a good place that explains it? I don't think ive been taught it yet but one of the listed possible solutions has arccosh in it, and im curious as to what exactly it is
     
  8. Feb 8, 2013 #7

    SammyS

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