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Math Help - Integral Question

  1. #1
    VkL
    VkL is offline
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    Integral Question

    How do you do this integral without a table?

    Integral [ Sqrt(4x^2 + 20)]

    I got it down to

    2* integral [ sqrt(x^2 + 5) ]

    Now I can use the table and get a answer that has a natural logarithm in it, But how do you do it without a table?? I know that's a stupid question to ask, But this does not look complicated to do!!! I just don't know where to start. U-sub does not work! ...by parts? I can't see how that would work..I was maybe thinking Switching it to polar coordinates?...Or maybe a trig substitution?

    Any help would be much appreciated!
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  2. #2
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    e^(i*pi)'s Avatar
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    From what I can see on Wolfram you should use a trig subsitution.

    Unfortunately I don't know anything about them
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  3. #3
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by VkL View Post
    How do you do this integral without a table?

    Integral [ Sqrt(4x^2 + 20)]

    I got it down to

    2* integral [ sqrt(x^2 + 5) ]

    Now I can use the table and get a answer that has a natural logarithm in it, But how do you do it without a table?? I know that's a stupid question to ask, But this does not look complicated to do!!! I just don't know where to start. U-sub does not work! ...by parts? I can't see how that would work..I was maybe thinking Switching it to polar coordinates?...Or maybe a trig substitution?

    Any help would be much appreciated!
    integral (1/(x^2+5)^(1/ 2)) dx - Wolfram|Alpha

    hit show steps

    or just remember those 12 or so table integrals and you will not have any problem you don't even have to learn them as must, because you will while practice learn them without knowing that you know them :P hehehehe

     \displaystyle \int \frac {dx}{\sqrt{x^2\pm a^2 } } = \ln |x+\sqrt{x^2 \pm a^2 }| + C
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  4. #4
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    Hello, VkL!

    Trig Substitution is the way to go . . . and it's still very messy!


    \displaystyle \int\!\sqrt{4x^2 + 20}\:dx \;=\;2\!\!\int\!\sqrt{x^2+5}\:dx

    Let: . x \,=\,\sqrt{5}\,\tan\theta \quad\Rightarrow\quad dx \:=\:\sqrt{5}\sec^2\!\theta\,d\theta

    And: . \sqrt{x^2+5} \:=\:\sqrt{5\tan^2\!\theta + 5} \:=\:\sqrt{5(\tan^2\!\theta+1)} \:=\:\sqrt{5\sec^2\!\theta} \:=\:\sqrt{5}\sec\theta


    Substitute: . \displaystyle 2\!\1\int(\sqrt{5}\sec\theta)(\sqrt{5}\sec^2\!\the  ta\,d\theta) \;=\;10\!\!\int\!\sec^3\!\theta\,d\theta


    This is the dreaded secant-cubed integral.
    . . It requires some tricky integration-by-parts.
    But I memorized the formula years ago . . .

    We have: . 10\cdot\frac{1}{2}\bigg[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg] + C


    Back-substitute: . \tan\theta \:=\:\dfrac{x}{\sqrt{5}} \:=\:\dfrac{opp}{adj}

    From Pythagorus: . opp = x,\;adj = \sqrt{5} \quad\Rightarrow\quad hyp = \sqrt{x^2+5}

    . . Hence: . \sec\theta = \dfrac{\sart{x^2+5}}{\sqrt{5}},\;\tan\theta = \dfrac{x}{\sqrt{5}}


    And we have: . 5\left[\dfrac{\sqrt{x^2+5}}{\sqrt{5}}\cdot\dfrac{x}{\sqrt  {5}} + \ln\left|\dfrac{\sqrt{x^2+5}}{\sqrt{5}} + \dfrac{x}{\sqrt{5}}\right|\, \right] + C

    . . . . . . . . =\;5\left[\dfrac{x\sqrt{x^2+5}}{5} + \ln\left|\dfrac{x+\sqrt{x^2+5}}{\sqrt{5}}\right|\,  \right] + C

    . . . . . . . . =\;x\sqrt{x^2+5} + 5\ln\left(\dfrac{x+\sqrt{x^2+5}}{\sqrt{5}}\right) + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The above answer is absolutely correct,
    . . but many textbooks simplify it beyond all recognition.


    We have:

    . . x\sqrt{x^2+5} + 5\bigg[\ln\left(x + \sqrt{x^2+5}\,\right) - \ln\left(\sqrt{5}\right)\,\bigg] + C

    . . =\;x\sqrt{x^2+5} + 5\ln\left(x+\sqrt{x^2+5}\,\right) - \underbrace{5\ln\left(\sqrt{5}\right)}_{\text{This is a constant!}} +\: C

    . . =\;x\sqrt{x^2+5} + 5\ln\left(x+\sqrt{x^2+5}\,\right) + C
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  5. #5
    MHF Contributor

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    I would look at \sqrt{4x^2+ 20}= 2\sqrt{x^2+ 5} and be reminded of " a^2+ b^2= c^2", the Pythagorean theorem. Or perhaps you will remember it better as " sin^2(\theta)+ cos^2(\theta)= 1", the trig identity derived from the Pythagorean theorem. Since that "5" is a constant and not a variable, divide the identity by cos^2(\theta) to get \frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)} or tan^2(\theta)+ 1= sec^2(\theta).

    That makes me think of the substitution x= \sqrt{5}tan(\theta) so that x^2+ 5= 5 tan^2(\theta)+ 5= 5(tan^2(\theta)+ 1)= 5 sec^2(\theta) and \sqrt{x^2 +5}= \sqrt{5} sec(\theta).

    Of course, if x= \sqrt{5}tan(\theta) then dx= \sqrt{5}sec^2(\theta)d\theta so we have
    2\int \sqrt{x^2+ 5} dx= 2\int \left(\sqrt{5}sec(\theta)\right)\left(\sqrt{5}sec^  2(\theta) d\theta= 10\int sec^3(\theta) d\theta = 10\int \frac{d\theta}{cos^3(\theta)}.

    Now that has cos(\theta) to an odd power so we can use a standard "trick"- multiply both numerator and denominator inside the integral by cos(x) to get
    10\int \frac{ cos(\theta) d\theta}{cos^4(\theta)}= 10\int \frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}

    Now let u= sin(\theta) so that du= cos(\theta)d\theta and the integral becomes
    10\int \frac{du}{(1- u^2)^2}

    The denominator factors as (1- u)^2(1+ u)^2 so "partial fractions" can be used to integrate it.
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