Well you can do
$$\int_0^{\pi}{\sin(\sin x)} = \pi H_0(1)\,,$$ where $H$ is a non-elementary function known as the Struve function. (For info on Struve functions, see http://mathworld.wolfram.com/StruveFunction.html.)
You might be wondering...what good does it do to express the integral in terms of a function that almost nobody has ever heard of?
Well, think back to a long time ago when the sine, cosine, and tangent functions themselves were as arcane as Struve functions are now. A math student might have asked a question analogous to your own, saying "Why hasn't anybody been able to integrate $\int{{dx\over 1+x^2}}$? The specialists at the time would have said, "Actually, we ${\it can}$ evaluate that integral:
$$\int{{dx\over 1+x^2}} = \arctan x\,.$$
If the student were to say, "What good does that do?" The expert would have said, well, this is helpful because see we have these books with values of the tangent function written down in them. So effectively we have "done the integral" because we know which book to look in.
Likewise today, it helps to express the integral $\int_0^{\pi}{\sin(\sin x)}$ as $\pi H_0(1)$ because now we know which book to look at. (Probably Abramowitz and Stegun's ${\it Handbook\ of\ Mathematical\ Functions}$, which I used to use a lot when I was a mathematical physicist.) More commonly, programs like Mathematica have all those values pre-loaded, along with approximation schemes for computing accurate values by interpolation.
And you know, that's all your calculator is doing when it computes $\sin x$ anyway. Once you're beyond the polynomial functions, you're pretty much in this territory.
$$\sin^4x\cos^2x=\left(\frac{1-\cos2x}2\right)^2\frac{1+\cos2x}2$$
$$=\frac{(1+\cos2x)(1-2\cos2x+\cos^22x)}8$$
$$=\frac{1 -\cos2x-\cos^22x+\cos^32x}8$$
Again, $\cos^22x=\dfrac{1+\cos4x}2$
and $\cos3y=4\cos^3y-3\cos y\iff 4\cos^3y=\cos3y+3\cos y$ set $y=2x$
Alternatively use Euler Identities
$$2\cos x= e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$$
$$(2i\sin x)^4(2\cos^2x)^2=(e^{ix}-e^{-ix})^4(e^{ix}+e^{-ix})^2$$
$$\implies64\sin^6x\cos^2x=(e^{ix}+e^{-ix})^2\cdot(e^{ix}+e^{-ix})^2(e^{ix}-e^{-ix})^2$$
$$=(e^{i2x}+e^{-i2x}+2)(e^{i2x}-e^{-i2x})^2$$
$$=(e^{i2x}+e^{-i2x}+2)(e^{i4x}+e^{-i4x}-2)$$
$$=e^{i6x}+e^{-i6x}-(e^{i2x}+e^{-i2x})+2(e^{i4x}+e^{-i4x})-4$$
$$=2\cos6x-(2\cos2x)+2(2\cos4x)-4$$
Best Answer
Since, $\frac{\sin(-x)}{1+(-x)^2}=-\frac{\sin(x)}{1+x^2}$ the function is symmetric over the interval of $[-1,1]$. Therefore we can evaluate the integral as:
Here is a visual representation of what I mean: