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Math Help - Trigonometric Substitution

  1. #1
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    Cool Trigonometric Substitution

    Hi everyone,

    I am a college student who is new to the math help forum. I would like to thank everyone in advance.

    Evaluate the indefinite integral:

    ∫ 36/(sqrt(35-18x-81x^2) dx

    I first manipulated the square root in the denominator:

    ∫sqrt(-81x^+18x-35)
    - ∫ sqrt( x^2-(2/9)x-(35/81) ) I removed the negative sign
    - ∫
    sqrt(x^2-2/9-35/81 +(1/81)) Complete the square
    - ∫ sqrt (x+(1/9)^2)

    So, (x + (1/9)) = 2/3 tan θ
    (3/2)x + 1/6= tan θ

    I then used u-substitution, where u = 2/3 tan θ and du= 2/3 sec^2θ dθ
    -∫ 36dθ/ u^2+(4/9)
    -36 ∫ (2/3) sec^2θ/ (2/3) secθ
    = -36∫ secθ
    =-36 | secθtanθ| + C
    θ= arctan (3/2)x+(1/6)

    I have typed this into my Webwork, and yet the solution is wrong...any suggestions are welcome.

    Also,
    I am unsure where to start for this problem:

    ∫e^(7x)/(e^(14x)+36)dx

    Would I use the natural log?

    Thanks so much in advance!

    -Nikayla
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by newman611 View Post
    Hi everyone,

    I am a college student who is new to the math help forum. I would like to thank everyone in advance.

    Evaluate the indefinite integral:

    ∫ 36/(sqrt(35-18x-81x^2) dx

    I first manipulated the square root in the denominator:

    ∫sqrt(-81x^+18x-35)
    - ∫ sqrt( x^2-(2/9)x-(35/81) ) I removed the negative sign
    Excuse me but how do you remove a negative sign inside a square root?
    \sqrt{-x}\ne -\sqrt{x}

    - ∫ sqrt(x^2-2/9-35/81 +(1/81)) Complete the square
    - ∫ sqrt (x+(1/9)^2)

    So, (x + (1/9)) = 2/3 tan θ
    (3/2)x + 1/6= tan θ

    I then used u-substitution, where u = 2/3 tan θ and du= 2/3 sec^2θ dθ
    -∫ 36dθ/ u^2+(4/9)
    -36 ∫ (2/3) sec^2θ/ (2/3) secθ
    = -36∫ secθ
    =-36 | secθtanθ| + C
    θ= arctan (3/2)x+(1/6)

    I have typed this into my Webwork, and yet the solution is wrong...any suggestions are welcome.

    Also,
    I am unsure where to start for this problem:

    ∫e^(7x)/(e^(14x)+36)dx

    Would I use the natural log?

    Thanks so much in advance!

    -Nikayla
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Mar 2010
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    For the second problem, you have \displaystyle \int \frac{e^{7x}}{e^{14x}+36}\;{dx} = \int \frac{e^{7x}}{(e^{7x})^2+36}\;{dx}.
    Let u = e^{7x} to get \displaystyle \frac{1}{7}\int\frac{1}{u^2+36}\;{du}. Let u = 6\tan{t} to get the answer.
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  4. #4
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    Thank you, HallsofIvy.
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  5. #5
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    Thanks, TheCoffeeMachine
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