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I do not figure out how to solve

$$ L= \frac{1}{2}\int_1^4 \sqrt{(16t^2+t+4)}\;dt $$

The key is probably in simplifying the polynomial equation, but I don't find a way that is really simplifying the integral. Can someone help?

Thank you.

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    $\begingroup$ Complete the square and use a trigonometric substitution. $\endgroup$
    – Mark Viola
    Dec 3, 2015 at 22:52
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    $\begingroup$ It's not that easy. One shall know also the method of Secant substitution. $\endgroup$
    – Enrico M.
    Dec 3, 2015 at 23:13
  • $\begingroup$ @Dr.MV The completion of the square was the first step to the resolution, but I did not go through the trigonometric substitution. I used a table of integrals that I could use. $\endgroup$
    – wayland700
    Dec 4, 2015 at 1:20
  • $\begingroup$ @KimPeekII Sure, it was more difficult that I thought, not knowing at first the table that was accessible. $\endgroup$
    – wayland700
    Dec 4, 2015 at 1:20
  • $\begingroup$ That's fine. Pleased to hear you have it now. $\endgroup$
    – Mark Viola
    Dec 4, 2015 at 1:22

2 Answers 2

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Notice, $$\frac{1}{2}\int_{1}^{4}\sqrt{16t^2+t+4}\ dt$$ $$=\frac{1}{2}\int_{1}^{4}4\sqrt{t^2+\frac{t}{16}+\frac{1}{4}}\ dt$$ $$=2\int_{1}^{4}\sqrt{\left(t+\frac{1}{32}\right)^2+\left(\frac{\sqrt {255}}{32}\right)^2}\ dt$$ using substitution, $t+\frac{1}{32}=t$, one should get $$=2\frac{1}{2}\left[\left(t+\frac{1}{32}\right)\sqrt{\left(t+\frac{1}{32}\right)^2+\left(\frac{\sqrt {255}}{32}\right)^2}+\frac{255}{32^2}\ln\left|\left(t+\frac{1}{32}\right)+\sqrt{\left(t+\frac{1}{32}\right)^2+\left(\frac{\sqrt {255}}{32}\right)^2}\right|\right]_1^4$$ I hope you can solve further

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  • $\begingroup$ Yes thank you, I found the formula in a book where there were tables of integrals and a section for those containing $\sqrt{a^2+u^2}, a >0$ For the one who is interested : $$ \int\sqrt{a^2+u^2} du = \frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}ln(u+\sqrt{a^2+u^2}) + C $$ $\endgroup$
    – wayland700
    Dec 4, 2015 at 1:12
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Not that easy. I will ignore the 1/2 factor and the extrema, by now.

$$16t^2 + t + 4 = \left(4t + \frac{1}{8}\right)^2 + \frac{255}{64}$$

Substitute then $z = 4t + \frac{1}{8}$ and $dz = 4 dt$ you have now:

$$\frac{1}{4}\int\sqrt{z^2 + \frac{255}{64}} dz$$

Now use the substitution $z = \frac{1}{8}\sqrt{255}\tan(s)$ and $dz = \frac{1}{8}\sqrt{255}\sec^2(s) dx$ this will lead to (make the math):

$$\frac{\sqrt{255}}{{32}}\int \frac{1}{8}\sqrt{255}\sec^3(s) ds = \frac{255}{256}\int \sec^3(s) dx$$

Now use the secant reduction formula:

$$\int \sec^m(x) dx = \frac{\sin(x)\sec^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int \sec^{m-2}(x) dx$$

and in your case $m = 3$. Remember that the integral of $\sec(x)$ is $\log(\tan(x) + \sec(x))$ and then you get:

$$\frac{255}{512}\tan(s)\sec(s) + \frac{255}{512}\log(\tan(s) + \sec(s))$$

Substitute back now. You had:

$$s = \arctan\left(\frac{8u}{255}\right)$$

and

$$u = \frac{1}{8} + 4y$$

obtaining:

$$\frac{1}{2}\left[\frac{1}{64}\sqrt{16t^2 + 4t + 4}(32t + 1) + \frac{255}{512}\log\left(\frac{8\sqrt{16t^2 + t + 4} + 32t + 1}{\sqrt{255}}\right)\right]_1^{4} = \text{do the math to obtain a number}$$

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  • $\begingroup$ Thank you for your answer. It is giving me the right result, but this trigonometric substitution was above the level of difficulty of my course. I found the formula to solve the integral in reference tables. The form of the equation you obtained is not identical to the reference formula, but they are similar in the way they use the expression with logarithms, but with different bases. $\endgroup$
    – wayland700
    Dec 4, 2015 at 1:35

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