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Math Help - Integrating 1 + 16x^2

  1. #1
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    Integrating 1 + 16x^2

    Evaluate the integration below:

    Integrating 1 + 16x^2-untitled.jpg
    Last edited by mr fantastic; February 16th 2011 at 11:13 PM.
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  2. #2
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    Have you tried a trig sub?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Have you tried a trig sub?
    yes i have tried it by using 4x = tan a but i m stuck at the last stage of integraiton, so plz help me
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  4. #4
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    What about \displaystyle x = \frac{1}{4}\tan u
    Last edited by pickslides; February 16th 2011 at 09:48 PM.
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  5. #5
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    Hello, soni!

    Your difficulty is understandable.


    \displaystyle \text{Evaluate: }\;\int^1_0\sqrt{1+16x^2}\,dx

    Let 4x \,=\,\tan\theta \quad\Rightarrow\quad dx \:=\:\frac{1}{4}\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{1+16x^2} \:=\:\sec\theta

    \displaystyle\text{Substitute: }\:\int \sec\theta\cdot\tfrac{1}{4}\sec^2\!\theta\,d\theta \;=\;\tfrac{1}{4}\int \sec^3\!\theta\,d\theta

    And we have the dreaded secant-cubed integral . . .


    The formula can be derived by integrating-by-parts twice.

    Or we can simply memorize the formula:
    . . \displaystyle \int\sec^3\!\alpha\,dx \:=\:\tfrac{1}{2}\bigg[\sec\alpha \tan\alpha + \ln\left|\sec\alpha + \tan\alpha\right|\bigg] + C


    So we have: . \frac{1}{4}\cdot\frac{1}{2}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C


    Back-substitute: . \tan\theta = 4x,\;\sec\theta = \sqrt{1+16x^2}

    And we have: . \frac{1}{8}\bigg[4x\sqrt{1+16x^2} + \ln\left|\sqrt{1+16x^2} + 4x\right|\bigg]^1_0


    Got it?

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  6. #6
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    What I would have done after reaching \frac{1}{4}\int sec^3(\theta)d\theta

    Rewrite it as \frac{1}4}\int\frac{d\theta}{cos^3(\theta)}

    Since that is now an odd power of cos(\theta), multiply both numerator and denominator by cos(\theta) to get

    \frac{1}{4}\int\frac{cos(\theta)d\theta}{cos^4(\th  eta)}= \frac{1}{4}\int\frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}
    and make the substitution u= sin(\theta):

    \frac{1}{4}\int\frac{du}{(1- u^2)^2}= \frac{1}{4}\int\frac{du}{(1- u)^2(1+ u)^2}
    a rational integral. It can be done by partial fractions:
    \frac{1}{(1- u)^2(1+ u)^2}= \frac{A}{1- u}+ \frac{B}{(1- u)^2}+ \frac{C}{1+ u}+ \frac{D}{(1+ u)^2}
    Eventually giving the same result Soroban got.
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