How much does symbolic integration mean to mathematics? (Before reading, I apologize for my poor English ability.)
I have enjoyed calculating some symbolic integrals as a hobby, and this has been one of the main source of my interest towards the vast world of mathematics. For instance, the integral below
$$ \int_0^{\frac{\pi}{2}} \arctan (1 - \sin^2 x \; \cos^2 x) \,\mathrm dx = \pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right). $$
is what I succeeded in calculating today.
But recently, as I learn advanced fields, it seems to me that symbolic integration is of no use for most fields in mathematics. For example, in analysis where the integration first stems from, now people seem to be interested only in performing numerical integration. One integrates in order to find an evolution of a compact hypersurface governed by mean curvature flow, to calculate a probabilistic outcome described by Ito integral, or something like that. Then numerical calculation will be quite adequate for those problems. But it seems that few people are interested in finding an exact value for a symbolic integral.
So this is my question: Is it true that problems related to symbolic integration have lost their attraction nowadays? Is there no such field that seriously deals with symbolic calculation (including integration, summation) anymore?
 A: I think it would be appropriate at this point to quote Forman Acton:

...at a more difficult but less pernicious level we have the
  inefficiencies engendered by exact analytic integrations where a sensible
  approximation would give a simpler and more effective algorithm. Thus
$$\begin{align*}\int_0^{0.3}\sin^8\theta\,\mathrm d\theta&=\left[\left(-\frac18\cos\,\theta\right)\left(\sin^4\theta+\frac76\sin^2\theta+\frac{35}{24}\right)\sin^3\theta+\frac{105}{384}\left(\theta-\sin\,2\theta\right)\right]_0^{0.3}\\ &=(-0.119417)(0.007627+0.101887+1.458333)(0.0258085)+0.004341\\ &=-0.0048320+0.0048341=0.0000021\end{align*}$$
manages to compute a very small result as the difference between two much
  larger numbers. The crudest approximation for $\sin\,\theta$ will give
$$\int_0^{0.3}\theta^8\,\mathrm d\theta=\frac19\left[\theta^9\right]_0^{0.3}=0.00000219$$
with considerably more potential accuracy and much less trouble. If
  several more figures are needed, a second term of the series may be kept.
In a similar vein, if not too many figures are required, the quadrature
$$\int_{0.45}^{0.55}\frac{\mathrm dx}{1+x^2}=\left[\tan^{-1}x\right]_{0.45}^{0.55}=0.502843-0.422854=0.079989\approx 0.0800$$
causes the computer to spend a lot of time evaluating two arctangents to get
  a result that would have been more expediently calculated as the product
  of the range of integration ($0.1$) by the value of the integrand at the
  midpoint ($0.8$). The expenditure of times for the two calculations is
  roughly ten to one. For more accurate quadrature, Simpson's rule would still
  be more efficient than the arctangent evaluations, nor would it lose a
  significant figure by subtraction. The student that worships at the altars of
  Classical Mathematics should really be warned that his rites frequently have
  quite oblique connections with the external world.

It may very well be that choosing the closed form approach would still end up with you having to (implicitly) perform a quadrature anyway; for instance, one efficient method for numerically evaluating the zeroth-order Bessel function of the first kind $J_0(x)$ uses the trapezoidal rule!
On the other hand, there are also situations where the closed form might be better for computational purposes. The usual examples are the complete elliptic integrals $K(m)$ and $E(m)$; both are more efficiently computed via the arithmetic-geometric mean than by using a numerical quadrature method.
But, as I said in the comments, for manipulational work, possessing a closed form for your integral is powerful stuff; there is a whole body of results that are now conveniently at your disposal once you have a closed form at hand. Think of it as "standing on the shoulders of giants".
In short, again, "it depends on the situation and the terrain".
A: I don't think your point of view is the right one. To compute an integral analytically and to compute an integral numerically are different things. A numerical analysis professor of mine once said that, in applications (engineering, physics...) it is often more convenient to directly evaluate integrals by numerical means, even if they are integrable analytically! For example, suppose that you need  
$$\int_{0}^{\frac{\pi}{2}} \arctan (1 - \sin^2 x \; \cos^2 x) \,\mathrm dx$$
meters of conducting wire. You make a phone call to the wire factory and ask for  what? For $\pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right)$ meters of wire? More realistically you will ask for something like $1.13$ meters of wire. 
To obtain this number $1.13$ you performed an approximation over the non-rational quantity $\pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right)$. In doing so you wasted information. It would have been more convenient (and, maybe, even more accurate) to perform this approximation on the first integral directly, that is, to evaluate it numerically.
Of course this does not render analytical methods useless. You could have a family of integrals depending on a parameter, for example. Numerical methods tell you nothing here. You could run across an integral in the middle of a proof, and need its exact value for theoretical purposes. The possibilities are countless.
A: If you're talking about practical engineering applications, then really only numerical approximations are used (and studied in computer science as 'numerical analysis' or more recently 'scientific computing'). 
As to an academic mathematical field nowadays that deals with symbolic integration, first some perspective. Newton/Leibniz invented integral calculus in ...hm...late 1600's and was popularized (as much as you can say that) in the 1700's. Some basic symbolic integration even occurred (without that name and system) before then. So let's just say there's been at least 300 years of work there.
Also, there's more to inverting derivatives than just integrals. Solving systems of partial differential equations seem to be the big thing (both numerically and symbolically) for almost as long as simple single variable integrals.
That said, there is a small academic group of people working in 'symbolic computation' (with their own journals), and one subarea is symbolic integration. There are proofs of impossibility (i.e. proving that given certain restrictions there is no 'closed form' for a particular integral), and there are algorithms for computing integrals given other certain restrictions (the Risch algorithm). The latter are often implemented in computer algebra packages (Mathematica, Maple, etc.).
There is surely room for solving particular integrals (in the AMM there don't seem to be many integrals though in the Problems section) and for finding patterns. I'd look at those journals to see what particular interest there is for integrals.
A: Symbolic integration becomes less popular indeed and most researchers prefer numerical integration. However, significance of the symbolic integration should not be underestimated. This can be shown by using the integration formula:
$$\int_0^1 f(x)dx = 2\sum_{m=1}^M{\sum_{n=0}^\infty{\frac{1}{{\left(2M\right)}^{2n+1}\left({2n+1}\right)!}\left.f^{(2n)}\left(x\right)\right|_{x=\frac{m-1/2}{M}}}}\,\,,$$
where the notation $f(x)^{(2n)}|_{x=\frac{m-1/2}{M}}$ implies 2n-th derivative at the points $x=\frac{m-1/2}{M}$. Once the integral is expanded as a series, we can use it either numerically or analytically (i.e. symbolically). It may be tedious to find by hand 2n-th derivatives. However, with powerful packages supporting symbolic programming like Maple, Mathematica or MATLAB this can be done easily. For example, by taking $f(x) = \frac{\theta}{1 + \theta^2 x^2}$ even at smallest $M = 1$ we can find a rapidly convergent series for the arctangent function:
$$\tan^{-1}(\theta)=i\sum_{n=1}^{\infty}\frac{1}{2n-1}\left(\frac{1}{\left(1+2i/\theta\right)^{2n-1}}-\frac{1}{\left(1-2i/\theta\right)^{2n-1}}\right),$$
where $i=\sqrt{-1}$.
This example shows that symbolic integration may be highly efficient in many numerical applications.
