Integrating Square Roots Containing Multiple Trigonometric Functions and/or Numbers When trying to calculate arc length of a curve I frequently come across problems that I do not know how to integrate, such as:
$$ \int{\sqrt{16\cos^2{4\theta} + \sin^2{4\theta}} d\theta} $$
Which in my attempt to solve I reduced to $ \int{\sqrt{17+15\cos{8\theta}\over 2} d\theta} $, or 
$$ \int{\sqrt{81\cos^2{3\theta} + 9\sin^2{3\theta} + 12\sin{3\theta} + 4}d\theta} $$
Which in my attempt to solve I reduced to $ \int{\sqrt{72\cos^2{3\theta} + 12\sin{3\theta} + 13}d\theta} $.
How would I go about solving these?
I have tried to use Wolfram's integral calculator to give me a clue as to how to solve these, but it was not able to provide me a useful answer.
This is from a calculus 3 university course.
 A: 
trying to calculate arc length of a curve I frequently come across

Computing the arc length of the $($co$)$sine function was one of the main historic reasons behind the discovery of elliptic integrals by Leonhard Euler over two centuries ago.
A: As others have noted this integral is very famous and belongs to the class of elliptic integrals (rightly named because they occur in evaluation of arc-length of an ellipse).  It is not possible to evaluate them in terms of elementary functions (exponential, logarithmic, trigonometric and algebraic functions), but there are formulas which can be used to evaluate them to any degree of accuracy without much hassle.
Let $$I(a, b) = \int_{0}^{\pi/2}\frac{dt}{\sqrt{a^{2}\cos^{2}t + b^{2}\sin^{2}t}}\tag{1}$$ be the elliptic integral of first kind where $a, b$ are positive real numbers.
If $a = b$ then $I(a, b) = \dfrac{\pi}{2a}$ (proof is almost obvious if we note that $\sin^{2}t + \cos^{2}t = 1$), but its bit difficult to handle the case when $a \neq b$. Gauss was smart enough to find a way to evaluate $I(a, b)$. Gausse proved that $$I(a, b) = I\left(\frac{a + b}{2}, \sqrt{ab}\right)\tag{2}$$ so that $a, b$ can be replaced respectively by their arithmetic and geometric means without affecting the value of integral $(1)$. It is not so difficult to prove that if we start with positive numbers $a, b$ and successively calculate their arithmetic and geometric means to develop the sequences $a_{n}, b_{n}$ such that $$a_{0} = a,\, b_{0} = b,\, a_{n + 1} = \frac{a_{n} + b_{n}}{2},\,b_{n + 1} = \sqrt{a_{n}b_{n}}\tag{3}$$ then both the sequences $a_{n}, b_{n}$ tend to a common limit which we denote by $M(a, b)$ and call the arithmetic-geometric mean of $a, b$. The great thing about this concept is that the convergence to the limit is very fast. Normally very few iterations are required to get $M(a, b)$ from $a, b$ (check using a calculator with say $a = 2, b = 5$). It is now obvious that $$I(a, b) = \frac{\pi}{2M(a, b)}\tag{4}$$ The integral in OP's post is called the elliptic integral of second kind namely $$J(a, b) = \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2}t + b^{2}\sin^{2}t}\,dt\tag{5}$$ and its evaluation is bit complicated. However we have the formula $$J(a, b) = \left(a^{2} - \frac{1}{2}\sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})\right)I(a, b)\tag{6}$$ The link between arithmetic-geometric mean and elliptic integrals was first noted by Gauss and this intrinsically beautiful theory is explained in detail and elementary manner in a series of posts on my blog.
A: I have a much simpler solution than the one previously proffered. Somebody is overthinking the problem. You say it's an arc length, so let consider an ellipse in the complex plane given by (we'll keep it quite general for now)
$$z=a\cos4\theta+ib\sin4\theta,\quad \theta\in[0,\pi/2]$$
The arc length is given by
$$s=\int_0^{\pi/2}|\dot z|~d\theta$$
Thus,
$$
\begin{align}
&\dot z=-4a\sin4\theta+i4b\cos4\theta\\
&|\dot z|=4\sqrt{a^2\sin^24\theta+b^2\cos^24\theta}=b\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^24\theta+1}\\
\end{align}
$$
We then find
$$
\begin{align}
s
&=\int_0^{\pi/2}b\sqrt{A\sin^24\theta+1}~d\theta,\quad A=\left(\frac{a^2-b^2}{b^2}\right)
\end{align}
$$
Now, let $x=4\theta$, then
$$
\begin{align}
s=
&b\int_0^{2\pi}\sqrt{A\sin^2x+1}~dx\\
&=4b\int_0^{\pi/2}\sqrt{A\sin^2x+1}~dx,\quad \text{by symmetry}\\
&=4b\text{E}(-A)=4b\frac{\pi}{2}~ _2\text{F}_1\left(-\frac 12,\frac 12;1;-A \right)
\end{align}
$$
where the elliptic integral is expressed in term of $\text{E}(k^2)$, as in Mathematica. The original integral in the OP corresponds to $a=1/4,~b=1$ and the arc length (circumference) of the ellipse is $s\approx 4.2892$. These results have been verified numerically for randomly chosen $a$ and $b$.
