Simple definite integral

Question

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.

Answer 1

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

Answer 2

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:

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