Integrating $\sin(\sqrt{16-x^2})$ with respect to x I am new to this site and not familiar with how to type out math notation so I will do my best. I have a problem I am working on regarding the volume of a circle wrapped around a cylinder of variable radius. For the first part of the problem I had no issue creating a function to represent the cross sectional area. Using this function and I am not trying to integrate to find the volume for different values of cylindrical radius r. The first half of the integral was painless but I have been stuck on the second half for a while now and am looking for some help as I cannot find a solution anywhere. Here is the integral;
$$\int_{-4}^4\sin(\sqrt{16-x^2})dx$$
This integral does include a couple other variable terms but they are treated as constants so there is little point in including them here as they will just complicate the problem. I am basically just looking for the technique used to integrate something like this because I am clueless. Thanks in advance.
 A: You can use trapezium rule
$$\int_{a}^b f(x) \ dx = \frac{1}{2}h(y_0+y_n)+2(y_1+y_2+y_3....y_{n-2}+y_{n-1})$$
Where $h=\frac{a+b}{n}$ and $y_n=f(x_n)$
A: I believe that this integral is non-elementary, i.e. it cannot be evaluated in terms of elementary functions. However, there are several numerical methods one may utilize to approximate the value of this integral. Indeed, as one answer mentions, one way would be through the use of the Trapezium Rule, which approximates the area under a curve using trapeziums. However, the Simpson's Rule may also be used. Basically, Simpson's Rule uses parabolas to approximate the area under a curve. The formula for Simpson's Rule (with $n$ equally spaced subdivisions over an interval) is given by:
$\int_{a}^{b}f(x)dx$ $\approx$ $\frac{h}{3} (f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+4f(x_n-1)+f(x_n))$
Here, $h=\frac{b-a}{n}$ and $x_i=a+ih$. Do note that, since every parabola is uniquely specified by $3$ points, Simpson's Rule requires an even number of intervals to be valid (and hence an odd number of ordinates). 
As an illustration of Simpson's Rule, we will use it to approximate the value of the proposed integral. Let $f(x)=sin(\sqrt{16-x^2})$. First observe that $\int_{-4}^{4} sin(\sqrt{16-x^2}) dx= 2 \int_{0}^{4} sin(\sqrt{16-x^2}) dx$, since the integrand is an odd function. Now, we can select 4 equally spaced subintervals, such that the ordinates are:
$x_0=0,x_1=1,x_2=2,x_3=3,x_4=4$
Thus, by applying the formula for Simpson's rule, we obtain the following approximation of the integral:
$2 \int_{0}^{4} f(x) dx \approx 2 \times \frac{1}{3} (f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4))$
Which can be easily found using a calculator. Of course, since the number of subintervals used is small, the answer produced may not be sufficiently accurate. Generally, to get more accurate approximations, simply use more subintervals.  
However, in this particular case, due to the shape of the curve, it just so happens to turn out that the Trapezium Rule provides a more accurate approximation to the integral (for the same number of intervals) as compared to the Simpson's rule, which is usually not the case.
