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Math Help - Inverse Trig Function Integrals

  1. #1
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    Inverse Trig Function Integrals

    I did this exercises, but I have a doubt in the process, thanks for your help.
    Attached Thumbnails Attached Thumbnails Inverse Trig Function Integrals-int.jpg  
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  2. #2
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    The integral,
    \int \frac{1}{x\sqrt{x^2-4}}dx
    Is almost the inverse cosecant integral.
    There are some adjustments. First,
    x^2-4=x^2-2^2
    Thus, it has form,
    \csc^{-1} (x/2)
    But note the following, that if the problem asked for,
    \int \frac{1}{|x|\sqrt{x^2-4}}dx
    Then the answer would be,
    \csc^{-1} (x/2)+C
    But it does not ask that.
    In that case the integral is,
    \csc^{-1} |x/2|+C
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  3. #3
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    Thanks but that is the integral of the secant, the cosecant is the same as the secant, but with negative sign. I did this:
    Attached Thumbnails Attached Thumbnails Inverse Trig Function Integrals-int2.jpg  
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  4. #4
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    Hello, ron007!

    If you're not familiar with the arcsecant formula,
    . . there's always Trig Substitution.


    \int\frac{dx}{x\sqrt{9x^2-16}}

    Let 3x \,=\,4\sec\theta\quad\Rightarrow\quad x \,=\,\frac{4}{3}\sec\theta\quad\Rightarrow\quad dx\,=\,\frac{4}{3}\sec\theta\tan\theta\,d\theta

    Then: . \sqrt{9x^2 - 16} \:=\:\sqrt{(3x)^2 - 16} \:=\:\sqrt{(4\sec\theta)^2 - 16} \:=\:\sqrt{16\sec^2\theta-16}

    . . =\:\sqrt{16(\sec^2\theta - 1)} \:=\:\sqrt{16\tan^2\theta} \:=\:4\tan\theta


    Substitute: . \int\frac{\frac{4}{3}\sec\theta\tan\theta\,d\theta  }{\frac{4}{3}\sec\theta\!\cdot\!4\tan\theta} \;=\;\frac{1}{4}\int d\theta \:=\:\frac{1}{4}\theta + C


    Back-substitute
    We have: . 3x \,=\,4\sec\theta\quad\Rightarrow\quad \sec\theta \,=\,\frac{3x}{4}\quad\Rightarrow\quad \theta\,=\,\text{arcsec}\frac{3x}{4}<br />


    Therefore, the answer is: . \frac{1}{4}\text{arcsec}\frac{3x}{4} + C

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, ron007!

    If you're not familiar with the arcsecant formula,
    . . there's always Trig Substitution.



    Let 3x \,=\,4\sec\theta\quad\Rightarrow\quad x \,=\,\frac{4}{3}\sec\theta\quad\Rightarrow\quad dx\,=\,\frac{4}{3}\sec\theta\tan\theta\,d\theta

    Then: . \sqrt{9x^2 - 16} \:=\:\sqrt{(3x)^2 - 16} \:=\:\sqrt{(4\sec\theta)^2 - 16} \:=\:\sqrt{16\sec^2\theta-16}

    . . =\:\sqrt{16(\sec^2\theta - 1)} \:=\:\sqrt{16\tan^2\theta} \:=\:4\tan\theta


    Substitute: . \int\frac{\frac{4}{3}\sec\theta\tan\theta\,d\theta  }{\frac{4}{3}\sec\theta\!\cdot\!4\tan\theta} \;=\;\frac{1}{4}\int d\theta \:=\:\frac{1}{4}\theta + C


    Back-substitute
    We have: . 3x \,=\,4\sec\theta\quad\Rightarrow\quad \sec\theta \,=\,\frac{3x}{4}\quad\Rightarrow\quad \theta\,=\,\text{arcsec}\frac{3x}{4}<br />


    Therefore, the answer is: . \frac{1}{4}\text{arcsec}\frac{3x}{4} + C

    I'm confussed .
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  6. #6
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    Quote Originally Posted by ron007 View Post
    I'm confussed .
    His explanation is very straightfoward. Have you done this kind of substitution? I hope your teacher didn't just ask you to memorize the formulas and not be able to derive them. What part are you confused on so he can help you more? Help us help you.
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  7. #7
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    Quote Originally Posted by Jameson View Post
    His explanation is very straightfoward. Have you done this kind of substitution? I hope your teacher didn't just ask you to memorize the formulas and not be able to derive them. What part are you confused on so he can help you more? Help us help you.
    I get the problem. I was doing an error in the substitution. Thank you guys for ur help.
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